{"id":29410,"date":"2022-03-02T17:15:24","date_gmt":"2022-03-02T11:45:24","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=29410"},"modified":"2022-03-02T17:15:24","modified_gmt":"2022-03-02T11:45:24","slug":"ap-10th-class-maths-bits-11th-lesson-in-telugu","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-10th-class-maths-bits-11th-lesson-in-telugu\/","title":{"rendered":"AP 10th Class Maths Bits 11th Lesson \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f"},"content":{"rendered":"

Practice the\u00a0AP 10th Class Maths Bits with Answers<\/a> 11th Lesson \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f on a regular basis so that you can attempt exams with utmost confidence.<\/p>\n

AP Board 10th Class Maths Bits 11th Lesson \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f<\/h2>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c281.
\n\"AP
\n\u0c2a\u0c48 \u0c2a\u0c1f\u0c02\u0c32\u0c4b sin C = \\(\\frac{3}{5}\\) \u0c05\u0c2f\u0c3f\u0c28 cos C \u0c35\u0c3f\u0c32\u0c41\u0c35. \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\"AP
\nsin C = \\(\\frac{3}{5}\\)
\nBC = 4 . 2.
\n(\u2235 3, 4, 5 \u0c2a\u0c4d\u0c30\u0c48\u0c25\u0c3e\u0c17\u0c30\u0c3f\u0c2f\u0c28\u0c4d\u200c \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c3e\u0c32\u0c41)
\n\u2234 cos C = \\(\\frac{\\mathrm{BC}}{\\mathrm{AC}}=\\frac{4}{5}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c282.
\nsin3<\/sup>\u03b8.cos \u03b8 + cos3<\/sup>\u03b8 . sin \u03b8 = sin \u03b8 cos \u03b8 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nL.H.S = sin3<\/sup>\u03b8 cos\u03b8 + cos3<\/sup>\u03b8 sin\u03b8
\n= sin \u03b8 cos \u03b8 (sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8)
\n= sin \u03b8 cos \u03b8 (1)
\n= sin \u03b8 cos \u03b8 = R.H.S.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c283.
\nsin \u03b8 = cos \u03b8 \u0c05\u0c2f\u0c3f\u0c28 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ? (0 < \u03b8 < 90)
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsin \u03b8 = cos \u03b8 \u0c05\u0c2f\u0c3f\u0c28
\n(cos 45 = sin 45 = \\(\\frac{1}{\\sqrt{2}}\\))
\n\u03b8= 45\u00b0
\n(\u0c32\u0c47\u0c26\u0c3e)
\nsin \u03b8 = cos \u03b8 \u21d2 cos (90 – \u03b8) = cos \u03b8
\n\u2234 90 – \u03b8 = 0 \u21d2 90 = 2\u03b8
\n\u21d2 \u03b8 = \\(\\frac{90}{2}\\) = 45\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c284.
\n\\(\\frac{1}{\\sec ^{2} A}+\\frac{1}{{cosec}^{2} A}\\) = 1 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nL.H.S = \\(\\frac{1}{\\sec ^{2} A}+\\frac{1}{{cosec}^{2} A}\\)
\n= cos2<\/sup> A + sin2<\/sup> A
\n= 1 = R.H.S.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c285.
\ncos 60\u00b0. cos 30\u00b0 + sin 60\u00b0 . sin 30\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ncos 60\u00b0 \u00b7 cos 30\u00b0 + sin 60\u00b0 : sin 30\u00b0
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c286.
\ntan \u03b8 cot \u03b8 = sec \u03b8 \u00b7x \u0c05\u0c2f\u0c3f\u0c28 x =
\nA) cos \u03b8
\nB) sec \u03b8
\nC) tan \u03b8
\nD) cot \u03b8
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nA) cos \u03b8<\/p>\n

tan \u03b8 cot \u03b8 = sec \u03b8 . x
\n1 = sec \u03b8 . x
\n\\(\\frac{1}{\\sec \\theta}\\) = x \u21d2 x = cos \u03b8.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c287.
\ncos 1\u00b0 . cos 2\u00b0 . cos 3o ….. cos 180\u00b0 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ncos 1\u00b0. cos 2\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c288.
\nx = sin \u03b8, y = cos \u03b8 \u0c05\u0c2f\u0c3f\u0c28 sin \u03b8, cos \u03b8 \u0c32 \u0c38\u0c30\u0c4d\u0c35\u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c3e\u0c28\u0c4d\u0c28\u0c3fx, y \u0c32\u0c32\u0c4b \u0c30\u0c3e\u0c2f\u0c17\u0c3e
\nA) x2<\/sup> – y2<\/sup> = 1
\nB) y2<\/sup> – x2<\/sup> = 1
\nC) x2<\/sup> + y2<\/sup> = 1
\nD) x2<\/sup> = y2<\/sup>.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nC) x2<\/sup> + y2<\/sup> = 1
\n[\u2235 sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1 :
\n\u21d2 x2<\/sup> + y2<\/sup> = 1]<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c289.
\ncos 12\u00b0 – sin 78\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ncos 12\u00b0 – sin 78\u00b0
\n= cos 12\u00b0 – sin (90\u00b0 – 12)\u00b0
\n= cos 12\u00b0 – cos 12\u00b0 = 0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2810.
\nx= cosec \u03b8 + cot \u03b8; y = cosec \u03b8 – cot \u03b8 \u0c05\u0c2f\u0c3f\u0c28 \u0c08 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c1f\u0c3f\u0c32\u0c4b \u0c38\u0c30\u0c48\u0c28\u0c26\u0c3f ………………….
\nA) x + y = 0
\nB) x – y = 0
\nC) \\(\\frac{x}{y}\\) = 1
\nD) xy = 1
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nD) xy = 1<\/p>\n

(cosec \u03b8 + cot \u03b8) (cosec \u03b8 – cot \u03b8).
\n= cosec2<\/sup> \u03b8 – cot2<\/sup> \u03b8 = 1
\n\u2234 xy = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2811.
\ncos (A – B) = ________
\nA) cos A cos B + sin A sin B
\nB) cos A sin A + cos B sin B
\nC) sin A sin B – cos A cos B
\nD) cos A cos B – sin A sin B
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nA) cos A cos B + sin A sin B<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2812.
\ncos (90\u00b0 – \u03b8) = ________
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ncos (90\u00b0 – \u03b8) = sin \u03b8<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2813.
\n\u2206ABC\u0c32\u0c4b of sin C = \\(\\frac{3}{5}\\) \u0c05\u0c2f\u0c3f\u0c28 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\"AP
\ncos A = \\(\\frac{\\mathrm{AB}}{\\mathrm{AC}}=\\frac{3}{5}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2814.
\ntan2<\/sup> \u03b8 – sec2<\/sup> \u03b8 = ________
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ntan2<\/sup> \u03b8 – sec2<\/sup> 0 = -(sec2<\/sup> \u2013 tan2<\/sup> A) = -1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2815.
\nsec (90\u00b0 – A) = ________
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsec (90\u00b0 – A) = cosec \u03b8<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2816.
\ncosec \u03b8 + cot \u03b8 = 5 \u0c05\u0c2f\u0c3f\u0c28 cosec \u03b8 – cot \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ncosec2<\/sup> \u03b8 – cot2<\/sup> \u03b8 = 1
\n(cosec \u03b8 + cot \u03b8) (cosec \u03b8 – cot \u03b8) = 1
\n5(cosec \u03b8 – cot \u03b8) = 1<\/p>\n

\u2234 (cosec \u03b8 – cot \u03b8) = \\(\\frac{1}{5}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2817.
\nx = 2 sec \u03b8; y = 2 tan \u03b8 \u0c05\u0c2f\u0c3f\u0c28 x2<\/sup> – y2<\/sup> = 4 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nx2<\/sup> – y2<\/sup> = (2 sec \u03b8)2<\/sup> \u2013 (2 tan \u03b8)2<\/sup>
\n= 4 sec2<\/sup> \u03b8 – 4 tan2<\/sup> \u03b8
\n= 4 (sec2<\/sup> \u03b8 – tan2<\/sup>\u03b8)
\n= 4 (1) = 4<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2818.
\n\u221a3 tan \u03b8 = 1 \u0c05\u0c2f\u0c3f\u0c28 9 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\u221a3 tan \u03b8 = 1 = tan \u03b8 = \\(\\frac{1}{\\sqrt{3}}\\)
\n\u2234 \u03b8 = 30\u00b0.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2819.
\n(sec 60\u00b0) (cos 60\u00b0) \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n(sec 60\u00b0) (cos 60\u00b0) = 1 (\u2235 sec \u03b8 – cos \u03b8 = 1)
\n(\u0c32\u0c47\u0c26\u0c3e)
\n(sec 60\u00b0) (cos 60\u00b0) = 2 \u00d7 \u00bd = 1.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2820.
\nsin (60\u00b0 + 30\u00b0) = ________
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsin (60\u00b0 + 30\u00b0) = sin 90\u00b0 = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2821.
\nsec \u03b8 + tan \u03b8 = 3 \u0c05\u0c2f\u0c3f\u0c28 sec \u03b8 – tan \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ? ________
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsec \u03b8 + tan \u03b8 = \u00bd
\n\u21d2 sec2<\/sup> \u03b8 – tan2<\/sup> \u03b8 = 1
\n\u21d2 (sec \u03b8 + tan \u03b8) (sec \u03b8 – tan \u03b8) = 1
\n\u21d2 \\(\\frac{1}{2}\\) (sec \u03b8 – tan \u03b8) = 1
\n\u2234 sec \u03b8 – tan \u03b8 = 1 \u00d7 2 = 2<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2822.
\ncot A = \\(\\frac{5}{12}\\) \u0c05\u0c2f\u0c3f\u0c28 sin A + cos A \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\"AP
\ncot A = \\(\\frac{5}{12}\\)
\nAC = 13 (\u2235 5, 12, 13 \u0c2a\u0c48\u0c25\u0c3e\u0c17\u0c30\u0c3f\u0c2f\u0c28\u0c4d \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c3e\u0c32\u0c41)
\n\u2234 sin A + cos A = \\(\\frac{12}{13}+\\frac{5}{13}=\\frac{17}{13}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2823.
\n\u0c08 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c1f\u0c3f\u0c32\u0c4b sin x \u0c15\u0c41 \u0c38\u0c3e\u0c27\u0c4d\u0c2f\u0c02 \u0c15\u0c3e\u0c28\u0c3f \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0f\u0c26\u0c3f?
\nA) \\(\\frac{3}{4}\\)
\nB) \\(\\frac{3}{5}\\)
\nC) \\(\\frac{4}{5}\\)
\nD) \\(\\frac{5}{4}\\)
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nD) \\(\\frac{5}{4}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2824.
\nsin \u03b8 = cos \u03b8 (0 < \u03b8 < 90) \u0c05\u0c2f\u0c3f\u0c24\u0c47 tan \u03b8 + cot \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsin \u03b8 = cos \u03b8
\n\u2234 \u03b8 = 45\u00b0. .
\n\u2234 tan \u03b8 + cot \u03b8 = tan 45 + cot 45
\n= 1 + 1 = 2
\n(\u0c32\u0c47\u0c26\u0c3e)
\nsin \u03b8 = cos \u03b8 \u21d2 \\(\\frac{\\sin \\theta}{\\cos \\theta}\\) = 1
\n\u21d2 tan \u03b8 =1
\n\u2234 cot \u03b8 = 1
\n\u0c15\u0c4a\u0c2c\u0c1f\u0c4d\u0c1f\u0c3f tan \u03b8 + cot \u03b8 = 1 + 1 = 2<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2825.
\nsec \u03b8 + tan \u03b8 = 3 \u0c05\u0c2f\u0c3f\u0c24\u0c47 sec \u03b8 – tan \u03b8 = \\(\\frac{1}{3}\\) \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsec \u03b8 + tan \u03b8 = 3
\n\u21d2 sec2<\/sup> \u03b8 – tan2<\/sup> \u03b8 = 1
\n\u21d2 (sec \u03b8 + tan \u03b8) (sec \u03b8 – tan \u03b8) = 1
\n\u21d2 3 (sec \u03b8 – tan \u03b8) = 1
\n\u2234 sec \u03b8 – tan \u03b8 = \\(\\frac{1}{3}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2826.
\n\u2206ABC \u0c28\u0c02\u0c26\u0c41 AB = c, BC = a, AC = b \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u2220BAC = \u03b8 \u0c05\u0c2f\u0c3f\u0c28 \u2206ABC \u0c35\u0c48\u0c36\u0c3e\u0c32\u0c4d\u0c2f\u0c2e\u0c41 ……….. (\u03b8 \u0c05\u0c32\u0c4d\u0c2a\u0c15\u0c4b\u0c23\u0c02 )
\nA) \\(\\frac{1}{2}\\) ab sin \u03b8
\nB) \\(\\frac{1}{2}\\) ca sin \u03b8
\nC) \\(\\frac{1}{2}\\) bc sin \u03b8
\nD) \\(\\frac{1}{2}\\) b2<\/sup> sin \u03b8
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nC) \\(\\frac{1}{2}\\) bc sin \u03b8<\/p>\n

\"AP
\n\u2206ADB \u0c32\u0c4b \u2220D = 90\u00b0
\nsin \u03b8 = \\(\\frac{\\mathrm{BD}}{\\mathrm{AB}}\\)
\n\u21d2 BD = AB sin \u03b8
\n= c sin \u03b8<\/p>\n

\u2206ABC \u0c35\u0c48\u0c36\u0c3e\u0c32\u0c4d\u0c2f\u0c02. = \\(\\frac{1}{2}\\) \u00d7 \u0c2d\u0c42\u0c2e\u0c3f \u00d7 \u0c0e\u0c24\u0c4d\u0c24\u0c41
\n= \\(\\frac{1}{2}\\) \u00d7 AC \u00d7 BD
\n= \\(\\frac{1}{2}\\) bc sin a<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2827.
\ntan \u03b8 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c35\u0c3f\u0c32\u0c41\u0c35 cosec \u03b8 \u0c2a\u0c26\u0c02\u0c32\u0c32\u0c4b \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ntan \u03b8 = \\(\\frac{1}{\\cot \\theta}=\\frac{1}{\\sqrt{{cosec}^{2} \\theta-1}}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2828.
\n\u0c08 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c1f\u0c3f\u0c28\u0c3f \u0c2a\u0c30\u0c3f\u0c36\u0c40\u0c32\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f :
\ni) sin2<\/sup> 20\u00b0 + sin2<\/sup> 70\u00b0 = 1
\nii) log2<\/sub> (sin 90\u00b0) = 1
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b \u0c38\u0c30\u0c3f\u0c2f\u0c48\u0c28\u0c26\u0c3f \u0c0f\u0c26\u0c3f ?
\nA) i) \u0c2e\u0c3e\u0c24\u0c4d\u0c30\u0c2e\u0c47
\nB) ii) \u0c2e\u0c3e\u0c24\u0c4d\u0c30\u0c2e\u0c47
\nC) i) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 ii) \u0c32\u0c41
\nD) i) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 ii) \u0c32\u0c41 \u0c15\u0c3e\u0c35\u0c41
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nA) i) \u0c2e\u0c3e\u0c24\u0c4d\u0c30\u0c2e\u0c47<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2829.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c0f \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c28\u0c3f\u0c30\u0c4d\u0c35\u0c1a\u0c3f\u0c02\u0c1a\u0c32\u0c47\u0c2e\u0c41 ?
\nA) sin 90\u00b0
\nB) cos 0\u00b0
\nC) sec 90\u00b0
\nD) cos 90\u00b0
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nC) sec 90\u00b0<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2830.
\n\\(\\sqrt{\\frac{1-\\cos ^{2} \\theta}{1+\\cot ^{2} \\theta}}\\) = sin2<\/sup> \u03b8 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\\(\\sqrt{\\frac{1-\\cos ^{2} \\theta}{1+\\cot ^{2} \\theta}}=\\sqrt{\\frac{\\sin ^{2} \\theta}{{cosec}^{2} \\theta}}=\\frac{\\sin \\theta}{{cosec} \\theta}\\)
\n= sin \u03b8 \u00b7 sin \u03b8 = sin2<\/sup> \u03b8.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2831.
\ntan 36\u00b0. tan 54\u00b0 + sin 30\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ntan 36\u00b0. tan 54\u00b0 + sin 30\u00b0
\n= tan 36\u00b0. tan (90\u00b0 – 36\u00b0) + \\(\\frac{1}{2}\\)
\n= tan 36\u00b0. cot 36\u00b0 + \\(\\frac{1}{2}\\) = 1 + \\(\\frac{1}{2}\\) = \\(\\frac{3}{2}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2832.
\nsin A = \\(\\frac{24}{25}\\) \u0c05\u0c2f\u0c3f\u0c28 sec A\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\"AP
\nsin A = \\(\\frac{24}{25}\\)
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n(25)2<\/sup> = AB2<\/sup> + (24)2<\/sup>
\n252<\/sup> \u2013 242<\/sup> = AB2<\/sup>
\n(25 + 24) (25 – 24) = AB2<\/sup>
\n49(1) = AB2<\/sup>
\nAB = \\(\\sqrt{49}\\) = 7
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2833.
\n4 cos2<\/sup> \u03b8 – 3 = 0 \u0c05\u0c2f\u0c3f\u0c28 sin \u03b8 = \\(\\frac{1}{2}\\) … \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n4 cos2<\/sup> \u03b8 – 3 = 0
\n\u21d24 cos2<\/sup> \u03b8 = 3
\n\u21d2 cos2<\/sup> \u03b8 = \\(\\frac{3}{4}\\) = cos \u03b8 = \\(\\sqrt{\\frac{3}{4}}=\\frac{\\sqrt{3}}{2}\\)
\n\u2234 \u03b8 = 30\u00b0
\n\u2234 sin \u03b8 = sin 30\u00b0 = \\(\\frac{1}{2}\\)
\n(\u0c32\u0c47\u0c26\u0c3e)
\n4 cos2<\/sup> \u03b8 – 3 = 0
\ncos2<\/sup> \u03b8 = \\(\\frac{3}{4}\\)
\nsin2<\/sup> \u03b8 = 1 – cos2<\/sup> \u03b8 = 1 – \\(\\frac{3}{4}\\)
\nsin2<\/sup> \u03b8 = \\(\\frac{4-3}{4}=\\frac{1}{4}\\)
\n:: sin \u03b8 = \\(\\sqrt{\\frac{1}{4}}=\\frac{1}{2}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2834.
\ncosec \u03b8 + cot \u03b8 = 2 \u0c05\u0c2f\u0c3f\u0c28 cos \u03b8 \u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2835.
\ncos (A + B) = 0 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 cos B = \\(\\frac{\\sqrt{3}}{2}\\) \u0c05\u0c2f\u0c3f\u0c28 ‘A’ \u0c15\u0c4b\u0c23\u0c2e\u0c41\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ncos (A + B) = 0 \u21d2 A + B = 90\u00b0
\ncos B = \\(\\frac{\\sqrt{3}}{2}\\) = B = 30\u00b0
\n\u21d2 A + B = 90\u00b0 \u21d2 A + 30\u00b0 = 90\u00b0
\n\u2234 A= 60\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2836.
\nsin x = \\(\\frac{5}{7}\\) \u0c05\u0c2f\u0c3f\u0c28 cosec x \u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsin x = \\(\\frac{5}{7}\\) \u0c05\u0c2f\u0c3f\u0c24\u0c47 cosec x = \\(\\frac{7}{5}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2837.
\n\u2220A = 75\u00b0, \u2220B = 30\u00b0, \u0c05\u0c2f\u0c3f\u0c28 tan (A – B) = 1 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ntan (A – B) = tan (75\u00b0 – 30\u00b0)
\n= tan 45\u00b0 = 1 .<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2838.
\nsec \u03b8 + tan \u03b8 = \\(\\frac{1}{3}\\) \u0c05\u0c2f\u0c3f\u0c28 sec \u03b8 – tan \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsec \u03b8 + tan \u03b8 = \\(\\frac{1}{3}\\)
\n\u21d2 sec2<\/sup> \u03b8 – tan2<\/sup> \u03b8 = 1
\n\u21d2 (sec \u03b8 + tan \u03b8) (sec \u03b8 – tan \u03b8) = 1
\n\u21d2 \\(\\frac{1}{3}\\) (sec \u03b8 – tan \u03b8) = 1
\n\u21d2 (sec \u03b8 – tan \u03b8) = 1 \u00d7 3
\n\u2234 sec \u03b8 – tan \u03b8 = 3<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2839.
\n\u0c12\u0c15 \u0c17\u0c21\u0c3f\u0c2f\u0c3e\u0c30\u0c02\u0c32\u0c4b 20 \u0c28\u0c3f\u0c2e\u0c3f\u0c37\u0c3e\u0c32 \u0c35\u0c4d\u0c2f\u0c35\u0c27\u0c3f\u0c32\u0c4b \u0c28\u0c3f\u0c2e\u0c3f\u0c37\u0c3e\u0c32 \u0c2e\u0c41\u0c32\u0c4d\u0c32\u0c41 \u0c1a\u0c47\u0c2f\u0c41 \u0c15\u0c4b\u0c23\u0c02 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\u0c17\u0c21\u0c3f\u0c2f\u0c3e\u0c30\u0c02\u0c32\u0c4b 20 \u0c28\u0c3f\u0c2e\u0c3f\u0c37\u0c3e\u0c32 \u0c35\u0c4d\u0c2f\u0c35\u0c27\u0c3f\u0c32\u0c4b \u0c28\u0c3f\u0c2e\u0c3f\u0c37\u0c3e\u0c32 \u0c2e\u0c41\u0c32\u0c4d\u0c32\u0c41 \u0c1a\u0c47\u0c2f\u0c41 \u0c15\u0c4b\u0c23\u0c2e\u0c41 = \\(\\frac{20}{60}\\) \u00d7 360\u00b0 = 120\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2840.
\nsec \u03b8 – tan \u03b8 = 3 \u0c05\u0c2f\u0c3f\u0c28 sec \u03b8 + tan \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsec \u03b8 – tan \u03b8 = 3
\n\u21d2 sec2<\/sup>\u03b8 – tan2<\/sup>\u03b8 = 1
\n\u21d2 (sec \u03b8 + tan \u03b8) (sec \u03b8 – tan \u03b8) = 1
\n\u21d2 sec \u03b8 + tan \u03b8 (3) = 1
\n\u2234 sec \u03b8 + tan \u03b8 = \\(\\frac{1}{3}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2841.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c2a\u0c1f\u0c2e\u0c41\u0c32\u0c4b BC \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\"AP
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ntan 30\u00b0 = AB
\n\u21d2 \\(\\frac{1}{\\sqrt{3}}=\\frac{7}{\\mathrm{BC}}\\) \u21d2 BC = 7\u221a3<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2842.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c2a\u0c1f\u0c02 \u2206ABC \u0c32\u0c4b AB = ________ \u0c38\u0c46\u0c02.\u0c2e\u0c40.
\n\"AP
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nAB2<\/sup> = AC2<\/sup> – BC2<\/sup>
\n= 172<\/sup> – 152<\/sup> = 289 – 225 = 64
\n\u2234 AB = \\(\\sqrt{64}\\) = 8.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2843.
\nsin 2\u03b8 = cos 3\u03b8 \u0c05\u0c2f\u0c3f\u0c28 \u03b8 = 15\u00b0 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsin 2\u03b8 = cos 3\u03b8
\nsin 2\u03b8 = sin (90 – 3\u03b8)
\n\u21d2 2\u03b8 = 90 – 3\u03b8
\n\u21d2 2\u03b8 + 3\u03b8 = 90\u00b0
\n\u21d2 5\u03b8 = 90\u00b0 \u21d2 \u03b8 = \\(\\frac{90^{\\circ}}{5}\\) = 18\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2844.
\ncos \u03b8 = \\(\\frac{3}{5}\\) \u0c05\u0c2f\u0c3f\u0c28 sin \u03b8 = ________
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ncos \u03b8 = \\(\\frac{3}{5}\\) \u0c05\u0c2f\u0c3f\u0c28 sin \u03b8 = \\(\\frac{4}{5}\\)
\n(\u22353, 4, 5 \u0c2a\u0c48\u0c25\u0c3e\u0c17\u0c30\u0c3f\u0c2f\u0c28\u0c4d \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c3e\u0c32\u0c41)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2845.
\ncos 60\u00b0 + sin 30\u00b0 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ncos 60\u00b0 + sin 30\u00b0 = \\(\\frac{1}{2}+\\frac{1}{2}\\) = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2846.
\nsec A + tan A = \\(\\frac{1}{5}\\) \u0c05\u0c2f\u0c3f\u0c28 sec A – tan A \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n5<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2847.
\ntan \u03b8 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c35\u0c4d\u0c2f\u0c41\u0c24\u0c4d\u0c24\u0c2e\u0c02\u0c28\u0c41 \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ncot \u03b8<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2848.
\n(sec2<\/sup>\u03b8 – 1) (cosec2<\/sup>\u03b8 – 1) = 1 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n(sec2<\/sup>\u03b8 \u2013 1) (cosec2<\/sup> \u03b8 – 1) = 1 .
\n= tan2<\/sup>\u03b8 – cot2<\/sup>\u03b8 = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2849.
\nsin (90 – A) = \\(\\frac{1}{2}\\) \u0c05\u0c2f\u0c3f\u0c24\u0c47 A \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n(90 – A) = \\(\\frac{1}{2}\\) \u21d2 90 – A = 30\u00b0
\n\u2234 90\u00b0 – 30\u00b0 = A \u21d2 A = 60\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2850.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b sin A = \\(\\frac{5}{13}\\) \u0c28\u0c15\u0c41 \u0c38\u0c30\u0c3f\u0c2a\u0c21\u0c41 \u0c2a\u0c1f\u0c02
\n\"AP
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2851.
\nsin2<\/sup> 60\u00b0 – sin2<\/sup> 30\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsin2<\/sup> 60\u00b0 – sin2<\/sup> 30\u00b0 = \\(\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}-\\left(\\frac{1}{2}\\right)^{2}\\)
\n= \\(\\frac{3}{4}-\\frac{1}{4}=\\frac{2}{4}=\\frac{1}{2}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2852.
\nsin \u03b8 = \\(\\frac{a}{b}\\), \u0c05\u0c2f\u0c3f\u0c28 tan \u03b8 \u0c28\u0c41 a, b \u0c32\u0c32\u0c4b \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsin \u03b8 = \\(\\frac{a}{b}\\)
\n\u2234 cos \u03b8 = \\(\\sqrt{1-\\sin ^{2} \\theta}\\)
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2853.
\ncos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2854.
\nsin \u03b8 = cos \u03b8 \u0c05\u0c2f\u0c3f\u0c28 2 tan \u03b8 + cos2<\/sup> \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsin \u03b8 = cos \u03b8 \u0c05\u0c2f\u0c3f\u0c28 \u03b8 = 45\u00b0
\n\u2234 2 tan \u03b8 + cos2 \u03b8 = 2 tan 45\u00b0 + cos2<\/sup> 45\u00b0
\n= 2(1) + \\(\\left(\\frac{1}{\\sqrt{2}}\\right)^{2}\\)
\n= 2 + \\(\\frac{1}{2}=\\frac{5}{2}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2855.
\ncosec \u03b8 = 2 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 cot \u03b8 = \u221a3p, \u0c07\u0c15\u0c4d\u0c15\u0c21 \u03b8 \u0c32\u0c18\u0c41\u0c15\u0c4b\u0c23\u0c02 \u0c05\u0c2f\u0c3f\u0c28 ‘p’ \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ncosec \u03b8 = 2, cot \u03b8 = \u221a3p
\ncosec2<\/sup> \u03b8 – cot2<\/sup> \u03b8 = 1.
\n= (2)2<\/sup> – (\u221a3p)2<\/sup> = 1
\n= 4 – 3p2<\/sup> = 1
\n= -3p2<\/sup> = 1 – 4 = -3 .
\n\u2234 p2<\/sup> = \\(\\frac{-3}{-3}\\) = 1
\n\u2234 p = \u221a1 = \u00b11<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2856.
\ntan \u03b8 + sec \u03b8 = 8 \u0c05\u0c2f\u0c3f\u0c28 sec \u03b8 – tan \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsec \u03b8 – tan \u03b8 = \\(\\frac{1}{8}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2857.
\n\\(\\left(\\frac{11}{\\cot ^{2} \\theta}-\\frac{11}{\\cos ^{2} \\theta}\\right)\\) \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\\(\\frac{11}{\\cot ^{2} \\theta}-\\frac{11}{\\cos ^{2} \\theta}\\) = 11\\(\\left(\\frac{1}{\\cot ^{2} \\theta}-\\frac{1}{\\cos ^{2} \\theta}\\right)\\)
\n= 11 [tan2<\/sup>\u03b8 – sec2<\/sup> \u03b8]
\n=-11 [sec2<\/sup> \u03b8 – tan2<\/sup>\u03b8]
\n= -11 (1) = -11<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2858.
\nsec 2A = cosec (A – 27\u00b0) \u0c05\u0c2f\u0c3f\u0c28 \u2220A \u0c35\u0c3f\u0c32\u0c41\u0c35 (2A \u0c32\u0c18\u0c15\u0c4b\u0c23\u0c2e\u0c41) \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsec 2A = cosec (A – 27\u00b0)
\ncosec (90 – 2A) = cosec (A – 27\u00b0)
\n\u2234 90 – 2A = A – 27
\n\u21d2 90 + 27 = A + 2A
\n\u21d2 3A = 117 = A = \\(\\frac{117}{3}\\) = 39\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2859.
\n\\(\\frac{1-\\sec ^{2} A}{{cosec}^{2} A-1}\\) = – tan4<\/sup> A \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nL.H.S. = \\(\\frac{1-\\sec ^{2} A}{{cosec}^{2} A-1}\\)
\n= \\(\\frac{-\\left(\\sec ^{2} A-1\\right)}{{cosec}^{2} A-1}=\\frac{-\\tan ^{2} A}{\\cot ^{2} A}\\)
\n= – tan2<\/sup> A X tan2<\/sup> A
\n= – tan4<\/sup> A = R.H.S.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2860.
\nsin \u03b8 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c17\u0c30\u0c3f\u0c37\u0c4d\u0c1f \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsin \u03b8 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c17\u0c30\u0c3f\u0c37\u0c4d\u0c1f \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2861.
\ntan \u03b8 = \\(\\frac{7}{8}\\) \u0c05\u0c2f\u0c3f\u0c28 \\(\\frac{(1+\\sin \\theta)(1-\\sin \\theta)}{(1+\\cos \\theta)(1-\\cos \\theta)}\\)
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2862.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b \u0c0f\u0c26\u0c3f \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c40\u0c2f \u0c38\u0c30\u0c4d\u0c35 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41 \u0c15\u0c3e\u0c26\u0c41?
\nA) sin2<\/sup> 0 + cos2<\/sup> 0 = 1
\nB) sec2<\/sup> 0 – tan2<\/sup> 0 = 1
\nC) sec2<\/sup> 0 + cosec2<\/sup> 0 = 1
\nD) cosec2<\/sup> 0 – cot2<\/sup> 0 = 1
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nC) sec2<\/sup> 0 + cosec2<\/sup> 0 = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2863.
\nsec \u03b8 = 3k \u0c2e\u0c30\u0c3f\u0c2f\u0c41 tan \u03b8 = \\(\\frac{3}{k}\\) \u0c05\u0c2f\u0c3f\u0c28 \\(\\left(\\mathbf{k}^{2}-\\frac{1}{\\mathbf{k}^{2}}\\right)=\\frac{1}{9}\\) \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsec \u03b8 = 3k, tan \u03b8 = \\(\\frac{3}{k}\\)
\nsec2<\/sup> \u03b8 – tan2<\/sup> \u03b8 = 1
\n\u21d2 (31)2<\/sup> – \\(\\left(\\frac{3}{\\mathrm{k}}\\right)^{2}\\) = 1
\n\u21d2 9\\(\\left(\\mathrm{k}^{2}-\\frac{1}{\\mathrm{k}^{2}}\\right)\\) = 1 \u21d2 k – \\(\\frac{1}{\\mathrm{k}^{2}}=\\frac{1}{9}\\)<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2864.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c28\u0c3f \u0c1c\u0c24\u0c2a\u0c30\u0c1a\u0c21\u0c02\u0c32\u0c4b \u0c38\u0c30\u0c48\u0c28 \u0c38\u0c2e\u0c3e\u0c27\u0c3e\u0c28\u0c3e\u0c28\u0c4d\u0c28\u0c3f<\/p>\n\n\n\n\n\n\n
i) sin 90\u00b0<\/td>\na) 0<\/td>\n<\/tr>\n
ii) cosec 45\u00b0<\/td>\nb) 1<\/td>\n<\/tr>\n
iii)tan 0\u00b0<\/td>\nc) \u221a2<\/td>\n<\/tr>\n
iv)cot 30\u00b0<\/td>\nd ) \u221a3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

A) i-b, ii-c, iii-a, iv-d
\nB) i-b, ii-a, iii-d, iv-c
\nC) i-a, ii-c, iii-b, iv-d
\nD) i-a, ii-b, iii-c, iv-d
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nA) i-b, ii-c, iii-a, iv-d<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2865.
\nAPQR \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c05\u0c02\u0c24\u0c30\u0c15\u0c4b\u0c23\u0c3e\u0c32\u0c41 P, Q \u0c2e\u0c30\u0c3f\u0c2f\u0c41. R \u0c05\u0c2f\u0c3f\u0c28 tan \\(\\)
\nA) sin \\(\\left(\\frac{\\mathrm{R}}{2}\\right)\\)
\nB) cot R
\nC) cot \\(\\left(\\frac{\\mathrm{R}}{2}\\right)\\)
\nD) tan \\(\\left(\\frac{\\mathrm{R}}{2}\\right)\\)
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nC) cot \\(\\left(\\frac{\\mathrm{R}}{2}\\right)\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2866.
\n\\(\\frac{2 \\tan 30^{\\circ}}{1+\\tan ^{2} 30^{\\circ}}\\) \u0c35\u0c3f\u0c32\u0c41\u0c35
\nA) sin 60\u00b0
\nB) cos 30\u00b0
\nC) tan 60\u00b0
\nD) A \u0c2e\u0c30\u0c3f\u0c2f\u0c41 B
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nD) A \u0c2e\u0c30\u0c3f\u0c2f\u0c41 B
\n\"AP
\n\u2234 sin 60\u00b0 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 cos 30\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2867.
\nsin (x – 20)\u00b0 = cos (3x – 10)\u00b0 \u0c05\u0c2f\u0c3f\u0c28 ‘x’ \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsin (x – 20)\u00b0 = cos (3x – 10)
\n[\u2235 sin A = cos B \u0c05\u0c2f\u0c3f\u0c24\u0c47 A + B = 90]
\n\u2234 x – 20 + 3x – 10 = 90
\n\u21d2 4x = 90\u00b0 + 30\u00b0 = 120\u00b0
\n\u21d2 x = \\(\\frac{120^{\\circ}}{4}\\) = 30\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2868.
\n\\(\\frac{1}{\\sec \\theta}\\) 0 \u2264 \u03b8 \u2264 90\u00b0 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c17\u0c30\u0c3f\u0c37\u0c4d\u0c20 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n2<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2869.
\ncos2<\/sup> 17\u00b0 – sin2<\/sup> 73\u00b0 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ncos2<\/sup> 17\u00b0 – sin2<\/sup> 73\u00b0
\n= cos2<\/sup> (90\u00b0 – 73\u00b0) – sin2<\/sup> 73\u00b0
\n= sin2<\/sup> 73\u00b0 – sin2<\/sup> 73\u00b0 = 0
\n(\u2235 cos (90 – \u03b8) = sin \u03b8)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2870.
\nA = 30\u00b0 \u0c05\u0c2f\u0c3f\u0c28 sin 2A \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nA = 30\u00b0 \u0c05\u0c2f\u0c3f\u0c28 sin 2A = sin 60\u00b0 = \\(\\frac{\\sqrt{3}}{2}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2871.
\nsin 45\u00b0 + cos45\u00b0 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u221a2 \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2872.
\n\u0c32\u0c02\u0c2c\u0c15\u0c4b\u0c23 \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c\u0c02 ABC \u0c32\u0c4b, \u0c32\u0c02\u0c2c\u0c15\u0c4b\u0c23\u0c02 ‘C’ \u0c35\u0c26\u0c4d\u0c26 \u0c15\u0c32\u0c26\u0c41. tan A = \\(\\frac{8}{15}\\) \u0c05\u0c2f\u0c3f\u0c28 cosec2<\/sup>A – 1 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ntan A = \\(\\frac{8}{15}\\) \u0c2e\u0c30\u0c3f\u0c2f\u0c41
\ncosec2<\/sup> – A – 1 = cot2<\/sup> A
\n= \\(\\frac{1}{\\tan ^{2} A}=\\frac{1}{\\left(\\frac{8}{15}\\right)^{2}}=\\frac{225}{64}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2873.
\n\\(\\frac{1}{2}\\)tan2<\/sup> 45\u00b0 = sin2<\/sup> A \u0c2e\u0c30\u0c3f\u0c2f\u0c41 ‘A’ \u0c32\u0c18\u0c41\u0c15\u0c4b\u0c23\u0c2e\u0c41 \u0c05\u0c2f\u0c3f\u0c28 A \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\\(\\frac{1}{2}\\)tan2<\/sup> 45\u00b0 = sin2<\/sup> A
\n\\(\\frac{1}{2}\\)(1)2<\/sup> = sin2<\/sup> A
\n\u21d2 \\(\\frac{1}{2}\\) = sin2<\/sup> A \u21d2 sin A = \\(\\frac{1}{\\sqrt{2}}\\)
\n\u2234 A = 45\u00b0 (\u2235 sin45\u00b0 = \\(\\frac{1}{\\sqrt{2}}\\))<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2874.
\n\u0c35\u0c3e\u0c15\u0c4d\u0c2f\u0c02 – I : \u2206ABC \u0c32\u0c4b \u2220A, \u2220B \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u2220C \u0c32\u0c41 \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c \u0c05\u0c02\u0c24\u0c30 \u0c15\u0c4b\u0c23\u0c3e\u0c32\u0c48\u0c28 sin \\(\\frac{A+B}{2}\\) = cos\\(\\frac{C}{2}\\)
\n\u0c35\u0c3e\u0c15\u0c4d\u0c2f\u0c02 – II : x = sin \u03b8, y = cos \u03b8 \u0c05\u0c2f\u0c3f\u0c28 x2<\/sup> + y2<\/sup> = 0.
\nA) I \u0c2e\u0c3e\u0c24\u0c4d\u0c30\u0c2e\u0c47 \u0c38\u0c24\u0c4d\u0c2f\u0c02
\nB) II \u0c2e\u0c3e\u0c24\u0c4d\u0c30\u0c2e\u0c47 \u0c38\u0c24\u0c4d\u0c2f\u0c02
\nC) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 II \u0c32\u0c41 \u0c30\u0c46\u0c02\u0c21\u0c42 \u0c05\u0c38\u0c24\u0c4d\u0c2f\u0c02
\nD) I \u0c2e\u0c30\u0c3f\u0c2f\u0c41 II \u0c32\u0c41 \u0c30\u0c46\u0c02\u0c21\u0c42 \u0c38\u0c24\u0c4d\u0c2f\u0c02
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nC) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 II \u0c32\u0c41 \u0c30\u0c46\u0c02\u0c21\u0c42 \u0c05\u0c38\u0c24\u0c4d\u0c2f\u0c02<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2875.
\ntan \u03b8 =1 \u0c05\u0c2f\u0c3f\u0c28 \\(\\frac{5 \\sin \\theta+4 \\cos \\theta}{5 \\sin \\theta-4 \\cos \\theta}\\) \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41. (\u03b8 \u0c05\u0c32\u0c4d\u0c2a\u0c15\u0c4b\u0c23\u0c2e\u0c41)
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ntan \u03b8 = 1 \u21d2 \u03b8 = 45\u00b0 (\u2235 tan 45\u00b0 = 1)
\n\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2876.
\ncos 2\u03b8 = sin 4\u03b8 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 2\u03b8 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 4\u03b8 \u0c32\u0c41 \u0c32\u0c18\u0c41\u0c15\u0c4b\u0c23\u0c3e\u0c32\u0c48\u0c28 ‘\u03b8’ \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ncos 2\u03b8 = sin 4\u03b8
\nsin (90 – 2\u03b8) = sin 4\u03b8
\n\u2234 90 – 2\u03b8 = 4\u03b8 \u21d2 90 = 4\u03b8 + 2\u03b8 = 6\u03b8
\n\u2234 \u03b8 = \\(\\frac{90^{\\circ}}{6}\\) = 15\u00b0 \u2234 \u03b8 = 15\u00b0
\n(\u0c32\u0c47\u0c26\u0c3e)
\n2\u03b8 + 4\u03b8 = 90\u00b0 (\u2235 sin A = cos B \u21d2 A + B = 90\u00b0)
\n6\u03b8 = 90\u00b0 \u21d2 \u03b8 = \\(\\frac{90^{\\circ}}{6}\\) = 15\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2877.
\n\u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b \u0c0f\u0c26\u0c3f \u0c05\u0c2f\u0c3f\u0c28\u0c2a\u0c4d\u0c2a\u0c41\u0c21\u0c41 tan \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c3f \u0c28\u0c3f\u0c30\u0c4d\u0c35\u0c1a\u0c3f\u0c02\u0c1a\u0c32\u0c47\u0c2e\u0c41 ?
\nA) 90\u00b0
\nB) 60\u00b0
\nC) 30\u00b0
\nD) 0\u00b0
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nA) 90\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2878.
\nsin x = cos x, 0 \u2264 x \u2264 90\u00b0 \u0c05\u0c2f\u0c3f\u0c28 x \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nx = 45\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2879.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b tanA = \\(\\frac{5}{12}\\) \u0c28\u0c41 \u0c2a\u0c4d\u0c30\u0c3e\u0c24\u0c3f\u0c28\u0c3f\u0c27\u0c4d\u0c2f\u0c2a\u0c30\u0c41\u0c1a\u0c41\u0c1f\u0c15\u0c41 \u0c38\u0c30\u0c48\u0c28 \u0c2a\u0c1f\u0c2e\u0c41.
\n\"AP
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nD) \u0c2a\u0c48\u0c35\u0c28\u0c4d\u0c28\u0c40<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2880.
\nsin45\u00b0 – cos 45\u00b0 + cos 60\u00b0 = tan \u03b8 \u0c05\u0c2f\u0c3f\u0c28 \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsin 45\u00b0 . cos 45\u00b0 + cos 60\u00b0 = tan \u03b8
\n\\(\\frac{1}{\\sqrt{2}} \\times \\frac{1}{\\sqrt{2}}+\\frac{1}{2}\\) = tan \u03b8
\n\\(\\frac{1}{2}+\\frac{1}{2}\\) = tan \u03b8 \u21d2 tan \u03b8 = 1
\n\u2234 \u03b8 = 45\u00b0<\/p>\n

\u2192 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c2a\u0c1f\u0c02 \u0c28\u0c41\u0c02\u0c21\u0c3f ON = x; PN = y; OP = r; \u2220PON = \u03b8 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u2220PNO = 90\u00b0 \u0c05\u0c2f\u0c3f\u0c28 81 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 82 \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28\u0c32\u0c15\u0c41 \u0c38\u0c2e\u0c3e\u0c27\u0c3e\u0c28\u0c3e\u0c32\u0c41 \u0c30\u0c3e\u0c2f\u0c02\u0c21\u0c3f.
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2881.
\ncos \u03b8 \u0c28\u0c41 x, r’ \u0c32\u0c32\u0c4b \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ncos \u03b8 = \\(\\frac{\\mathbf{x}}{\\mathbf{r}}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2882.
\ntan \u03b8 \u0c28\u0c41 x, y \u0c32\u0c32\u0c4b \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ntan \u03b8 = \\(\\frac{\\mathbf{y}}{\\mathbf{r}}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2883.
\nsin \u03b8 cosec \u03b8 + cos \u03b8 . sec \u03b8 + tan \u03b8. cot \u03b8 = 3 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsin \u03b8 cosec \u03b8 + cos \u03b8 . sec \u03b8 + tan \u03b8. cot \u03b8
\n\"AP
\n= 1 + 1 + 1 = 3<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2884.
\nsin \u03b8. cosec \u03b8 = x \u0c05\u0c2f\u0c3f\u0c28 x \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsin \u03b8. cosec \u03b8 = x
\n\"AP
\n\u2234 x = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2885.
\nsin \u03b8 = \\(\\frac{a}{b}\\) \u0c05\u0c2f\u0c3f\u0c28 cos \u03b8\u0c28\u0c41 a, b \u0c32\u0c32\u0c4b \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c02\u0c21\u0c3f
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2886.
\nsin \u03b8 = 12 \u0c05\u0c2f\u0c3f\u0c28 tan \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2887.
\nsin \u03b8. sec \u03b8 = tan \u03b8 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nL.H.S. = sin \u03b8. sec \u03b8
\n= sin 8 . \\(\\frac{1}{\\cos \\theta}\\)
\n= \\(\\frac{\\sin \\theta}{\\cos \\theta}\\) = tan \u03b8 = R.H.S.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2888.
\n\\(\\sqrt{1+\\cot ^{2} \\theta}\\) = cosec \u03b8 \u0c05\u0c28\u0c3f \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c3f\u0c02\u0c1a\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\\(\\sqrt{1+\\cot ^{2} \\theta}=\\sqrt{{cosec}^{2} \\theta}\\) = cosec \u03b8<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2889.
\nsec \u03b8 = \\(\\frac{13}{12}\\) \u0c05\u0c2f\u0c3f\u0c28 sin \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2890.
\n\u0c1c\u0c24\u0c2a\u0c30\u0c41\u0c1a\u0c41\u0c2e\u0c41.
\n\"AP
\nA) i-c, ii-a, iii-d, iv-b
\nB) i-b, ii-d, iii-a, iv-c
\nC) i-a, ii-d, iii-c, iv-b
\nD) i-d, ii-b, iii-a, iv-c
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nB) i-b, ii-d, iii-a, iv-c<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2891.
\n\\(\\sqrt{1+\\sin A} \\cdot \\sqrt{1-\\sin A}\\) = cos A \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2892.
\n\u0c2a\u0c4d\u0c30\u0c35\u0c1a\u0c28\u0c02 – A: sin 0\u00b0, sin 30\u00b0, sin 90\u00b0 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c41 \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b \u0c15\u0c32\u0c35\u0c41.
\n\u0c2a\u0c4d\u0c30\u0c35\u0c1a\u0c28\u0c02 – B: tan 45\u00b0, sec 60\u00b0, cosec-30\u00b0 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c41 \u0c12\u0c15 \u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c28\u0c3f \u0c0f\u0c30\u0c4d\u0c2a\u0c30\u0c41\u0c38\u0c4d\u0c24\u0c3e\u0c2f\u0c3f.
\nA) A \u0c38\u0c24\u0c4d\u0c2f\u0c02, B \u0c05\u0c38\u0c24\u0c4d\u0c2f\u0c02
\nB) A \u0c05\u0c38\u0c24\u0c4d\u0c2f\u0c02, B \u0c38\u0c24\u0c4d\u0c2f\u0c02
\nC) A \u0c38\u0c24\u0c4d\u0c2f\u0c02, B \u0c38\u0c24\u0c4d\u0c2f\u0c02
\nD) A \u0c05\u0c38\u0c24\u0c4d\u0c2f\u0c02, B \u0c05\u0c38\u0c24\u0c4d\u0c2f\u0c02
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nC) A \u0c38\u0c24\u0c4d\u0c2f\u0c02, B \u0c38\u0c24\u0c4d\u0c2f\u0c02<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2893.
\ntan2<\/sup> 30\u00b0 + 2cot2<\/sup> 60\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ntan2<\/sup> 30\u00b0 + 2cot2<\/sup> 60\u00b0
\n= \\(\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}+2\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}\\)
\n= \\(\\frac{1}{3}+2 \\cdot \\frac{1}{3}=\\frac{1}{3}+\\frac{2}{3}=\\frac{3}{3}\\) = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2894.
\nsin2<\/sup> 75\u00b0 + cos2<\/sup> 75\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n1.
\n(sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2895.
\nsin4<\/sup> \u03b8 – cos4<\/sup> \u03b8 =
\nA) 1
\nB) cos2<\/sup> \u03b8 – sin2<\/sup> \u03b8
\nC) 2 sin2<\/sup> \u03b8 – 1
\nD) 2 sin2<\/sup> \u03b8
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nC) 2 sin2<\/sup> \u03b8 – 1<\/p>\n

sin4<\/sup> \u03b8 – cos4<\/sup> \u03b8 = (sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8) (sin2<\/sup> \u03b8 – cos2<\/sup> \u03b8)
\n= 1 (sin2<\/sup> \u03b8 – cos2<\/sup> \u03b8)
\n= sin2<\/sup> \u03b8 – cos2<\/sup> \u03b8
\n= sin2<\/sup> \u03b8 – (1 – sin2<\/sup> \u03b8)
\n= sin2<\/sup> \u03b8 – 1 + sin2<\/sup> \u03b8
\n= 2 sin2<\/sup> \u03b8 – 1<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2896.
\ntan \u03b8 = \\(\\frac{1}{\\sqrt{3}}\\); \u0c05\u0c2f\u0c3f\u0c28 cos \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ? (\u03b8 \u0c05\u0c32\u0c4d\u0c2a\u0c15\u0c4a\u0c23\u0c2e\u0c41)
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ntan \u03b8 = \\(\\frac{1}{\\sqrt{3}}\\) \u21d2 \u03b8 = 30\u00b0
\n\u2234 cos \u03b8 = cos 30\u00b0 = \\(\\frac{\\sqrt{3}}{2}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2897.
\n(1 + tan \u03b8)2<\/sup> = sec2<\/sup> \u03b8 + 2 tan \u03b8 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nL.H.S. = (1 + tan\u03b8)2<\/sup>
\n= 12<\/sup> + tan2<\/sup> \u03b8 + 2(1) tan \u03b8
\n(: (a + b)2<\/sup> = a2<\/sup> + b2<\/sup> + 2ab)
\n= 1 + tan2<\/sup> \u03b8 + 2 tan \u03b8
\n= sec2<\/sup> \u03b8 + 2 tan \u03b8 = R.H.S.
\nL.H.S. = R.H.S.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2898.
\ntan \u03b8 \u0c28\u0c41 sec \u03b8 \u0c32\u0c32\u0c4b \u0c24\u0c46\u0c32\u0c41\u0c2a\u0c17\u0c3e
\nA) \\(\\frac{\\sqrt{\\sec ^{2} \\theta-1}}{\\sec \\theta}\\)
\nB) \\(\\frac{\\sec \\theta}{\\sqrt{\\sec ^{2} \\theta-1}}\\)
\nC) \\(\\frac{1}{\\sqrt{\\sec ^{2} \\theta-1}}\\)
\nD) \\(\\sqrt{\\sec ^{2} \\theta-1}\\)
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nD) \\(\\sqrt{\\sec ^{2} \\theta-1}\\)<\/p>\n

sec2<\/sup> \u03b8 – tan2<\/sup> \u03b8 = 1
\nsec2<\/sup> A – 1= tan2<\/sup> \u03b8
\ntan \u03b8 = \\(\\sqrt{\\sec ^{2} \\theta-1}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2899.
\n5 sin A = 3 \u0c05\u0c2f\u0c3f\u0c28 sec2<\/sup> A – tan2<\/sup> A \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n5 sin A = 3 \u0c05\u0c2f\u0c3f\u0c28 sec2<\/sup> A – tan2<\/sup> A = 1
\n(\u2235 sec2<\/sup> A – tan2<\/sup> A = 1)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28100.
\ntan \u03b8 \u0c28\u0c41 sin \u03b8 \u0c32\u0c4b \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ntan \u03b8 = \\(\\frac{\\sin \\theta}{\\cos \\theta}=\\frac{\\sin \\theta}{\\sqrt{1-\\sin ^{2} \\theta}}\\)
\n(\u2235 sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1
\ncos2<\/sup> \u03b8 = 1 – sin2<\/sup> \u03b8
\ncos \u03b8 = \\(\\sqrt{1-\\sin ^{2} \\theta}\\))<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28101.
\ncos \u03b8 = \\(\\frac{\\sqrt{3}}{2}\\) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 8 \u0c05\u0c32\u0c4d\u0c2a\u0c15\u0c4b\u0c23\u0c2e\u0c41 \u0c05\u0c2f\u0c3f\u0c28 4 sin2<\/sup> \u03b8 + tan2<\/sup> \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ncos \u03b8 = \\(\\frac{\\sqrt{3}}{2}\\)
\n\u2234 \u03b8 = 30\u00b0
\n\u2234 4 sin2<\/sup> \u03b8 + tan2<\/sup> \u03b8
\n= 4 sin2<\/sup> 30\u00b0 + tan2<\/sup> 30\u00b0
\n= 4\\(\\left(\\frac{1}{2}\\right)^{2}\\) + \\(\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}\\) = 1 + \\(\\frac{1}{3}=\\frac{4}{3}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28102.
\n\"AP
\n\u0c2a\u0c1f\u0c02 \u0c28\u0c41\u0c02\u0c21\u0c3f, \u2206ABC \u0c32\u0c4b \u2220B = 90\u00b0; \u2220C = \u03b8 \u0c05\u0c2f\u0c3f\u0c28 tan \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c30\u0c3e\u0c2f\u0c02\u0c21\u0c3f.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ntan \u03b8 = \\(\\frac{\\mathrm{AB}}{\\mathrm{BC}}=\\frac{15}{8}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28103.
\n\"AP
\n‘\u0c2a\u0c1f\u0c02 \u0c28\u0c41\u0c02\u0c21\u0c3f cos \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ncos \u03b8 = \\(\\frac{\\mathrm{BC}}{\\mathrm{AC}}=\\frac{8}{17}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28104.
\ncos 0\u00b0 + sin 90\u00b0 + \u221a2 sin 45\u00b0 = 3 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ncos 0\u00b0 + sin 90\u00b0 + \u221a2 sin 45\u00b0
\n= 1 + 1 + \u221a2 \u00d7 \\(\\frac{1}{\\sqrt{2}}\\) = 1 + 1 + 1 = 3<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28105.
\n3 sin2<\/sup> 45\u00b0 + 2 cos2<\/sup> 60\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n3 sin2<\/sup> 45\u00b0 + 2 cos2<\/sup> 60\u00b0
\n= 3\\(\\left(\\frac{1}{\\sqrt{2}}\\right)^{2}\\) + 2\\(\\left(\\frac{1}{2}\\right)^{2}\\)
\n= \\(\\frac{3}{2}+2 \\times \\frac{1}{4}=\\frac{3}{2}+\\frac{1}{2}\\) = 2<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28106.
\nsin (log2<\/sub> 1)\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsin (log2<\/sub> 1)\u00b0 = sin 0\u00b0 = 0 (\u2235 log 1 = 0)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28107.
\na4<\/sup> – b4<\/sup> = (a2<\/sup> + b2<\/sup>) (a2<\/sup> – b2<\/sup>) \u0c05\u0c28\u0c47 \u0c38\u0c42\u0c24\u0c4d\u0c30\u0c3e\u0c28\u0c4d\u0c28\u0c3f \u0c09\u0c2a\u0c2f\u0c4b\u0c17\u0c3f\u0c02\u0c1a\u0c3f sin4<\/sup> \u03b8 – cos4<\/sup> \u03b8 \u0c28\u0c41 \u0c38\u0c42\u0c15\u0c4d\u0c37\u0c4d\u0c2e\u0c40\u0c15\u0c30\u0c3f\u0c02\u0c1a\u0c17\u0c3e \u0c35\u0c1a\u0c4d\u0c1a\u0c41 \u0c2b\u0c32\u0c3f\u0c24\u0c02 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b \u0c26\u0c47\u0c28\u0c3f\u0c15\u0c3f \u0c38\u0c2e\u0c3e\u0c28\u0c02 ?
\nA) sin2<\/sup>\u03b8 – cos2<\/sup>\u03b8
\nB) 2 sin2<\/sup>\u03b8 – 1
\nC) 1 – 2cos2<\/sup>\u03b8
\nD) \u0c2a\u0c48\u0c35\u0c28\u0c4d\u0c28\u0c40
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nD) \u0c2a\u0c48\u0c35\u0c28\u0c4d\u0c28\u0c40<\/p>\n

sin4<\/sup>\u03b8 – cos4<\/sup>\u03b8
\n= (sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8) (sin2<\/sup> \u03b8 – cos2<\/sup> \u03b8)
\n= sin2<\/sup> \u03b8 – cos2<\/sup> \u03b8 …….. (1)
\n= sin\u03b8 – (1 – sin2<\/sup>\u03b8)
\n= sin2<\/sup>\u03b8 – 1 + sin2<\/sup>\u03b8
\n= 2 sin2<\/sup>\u03b8 – 1 …….. (2)
\n(1) \u21d2 (1 – cos2<\/sup> \u03b8) – cos2<\/sup>\u03b8 = 1 – 2cos2<\/sup>\u03b8<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28108.
\ntan \u03b8 + cot \u03b8 = 2 \u0c05\u0c2f\u0c3f\u0c28 tan \u03b8 + cot \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ntan \u03b8 + cot \u03b8 = 2
\n(tan \u03b8 + cot \u03b8)2<\/sup> = (2)2<\/sup>
\n\"AP
\n= tan2<\/sup> \u03b8 + cot2<\/sup> \u03b8 + 2 = 4
\n= tan2<\/sup> \u03b8 + cot2<\/sup> \u03b8 = 4 – 2 = 2<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28109.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c2a\u0c1f\u0c4d\u0c1f\u0c3f\u0c15\u0c32\u0c4b \u0c32\u0c4b\u0c2a\u0c3f\u0c02\u0c1a\u0c3f\u0c28 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c02\u0c21\u0c3f.
\n\"AP
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\u221a3<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28110.
\n\u2206ABC \u0c32\u0c4b sin A = \\(\\frac{3}{5}\\), cos A = \\(\\frac{4}{5}\\) \u0c05\u0c2f\u0c3f\u0c28 \u0c06 \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c \u0c15\u0c4a\u0c32\u0c24\u0c32\u0c41 \u0c35\u0c30\u0c41\u0c38\u0c17\u0c3e BC, AB, AC \u0c32\u0c41 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b \u0c0f\u0c35\u0c3f \u0c15\u0c3e\u0c35\u0c1a\u0c4d\u0c1a\u0c41\u0c28\u0c41 ?
\n(A) 9, 12, 15
\n(B) 3, 4, 5,
\n(C) 12, 16, 20
\n(D) \u0c2a\u0c48\u0c35\u0c28\u0c4d\u0c28\u0c40
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41:
\nD) \u0c2a\u0c48\u0c35\u0c28\u0c4d\u0c28\u0c40<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28111.
\n(1 + tan2<\/sup> 60\u00b0)2<\/sup> \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n(1 + tan2<\/sup>60\u00b0)2<\/sup> = (1 + (\u221a3)2<\/sup>)2<\/sup>
\n= 42<\/sup> = 16<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28112.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c28\u0c3f \u0c1c\u0c24\u0c2a\u0c30\u0c41\u0c1a\u0c41\u0c2e\u0c41.<\/p>\n\n\n\n\n\n
i) sec \u03b8<\/td>\na) \\(\\frac{1}{\\sin \\theta}\\)<\/td>\n<\/tr>\n
ii) cosec \u03b8<\/td>\nb) \\(\\frac{1}{\\tan \\theta}\\)<\/td>\n<\/tr>\n
iii)cot \u03b8<\/td>\nc) \\(\\frac{1}{\\cos \\theta}\\)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

A) i-a, ii-b, iii-c
\nB) i-c, ii-a, iii-b
\nC) i-b, ii-a, iii-c,
\nD) i-a, ii-c, iii-b
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nB) i-c, ii-a, iii-b<\/p>\n

\u2192 \u0c07\u0c35\u0c4d\u0c35\u0c2c\u0c21\u0c3f\u0c28 \u0c2a\u0c1f\u0c3e\u0c28\u0c4d\u0c28\u0c3f \u0c2a\u0c30\u0c3f\u0c36\u0c40\u0c32\u0c3f\u0c02\u0c1a\u0c3f, 113, 114 \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28\u0c32\u0c15\u0c41 \u0c1c\u0c35\u0c3e\u0c2c\u0c41\u0c32\u0c41 \u0c30\u0c3e\u0c2f\u0c02\u0c21\u0c3f.
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28113.
\nsin A \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nsin A = \\(\\frac{12}{13}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28114.
\ntan C \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c30\u0c3e\u0c2f\u0c02\u0c21\u0c3f.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ntan C = \\(\\frac{5}{12}\\)<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28115.
\n\u03b1 + \u03b2 = 90\u00b0 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u03b1 = 2\u03b2 \u0c05\u0c2f\u0c3f\u0c28 cos2<\/sup> \u03b1 + sin2<\/sup> \u03b2 = \\(\\frac{1}{2}\\) \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\u03b1 + \u03b2 = 90\u00b0, \u03b1 = 2\u03b2
\n2\u03b2 + \u03b2 = 90\u00b0 \u21d2 3\u03b2 = 90\u00b0 \u21d2 \u03b2 = 30\u00b0
\n\u2234 \u03b1 = 60\u00b0
\ncos2<\/sup> \u03b1 + sin2<\/sup> \u03b2 = cos2<\/sup> 60\u00b0 + sin2<\/sup> 30\u00b0
\n= \\(\\left(\\frac{1}{2}\\right)^{2}+\\left(\\frac{1}{2}\\right)^{2}=\\frac{1}{4}+\\frac{1}{4}=\\frac{2}{4}=\\frac{1}{2}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28116.
\n\u2206ABC \u0c32\u0c02\u0c2c\u0c15\u0c4b\u0c23 \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c\u0c02\u0c32\u0c4b \u2220B = 90\u00b0, BC = 24 \u0c38\u0c46\u0c02.\u0c2e\u0c40. \u0c2e\u0c30\u0c3f\u0c2f\u0c41 tan C = \\(\\frac{5}{12}\\) \u0c05\u0c2f\u0c3f\u0c28 \u0c15\u0c30\u0c4d\u0c23\u0c2e\u0c41 AC \u0c2a\u0c4a\u0c21\u0c35\u0c41 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\"AP
\nBC = 24 \u0c38\u0c46\u0c02.\u0c2e\u0c40. \u0c2e\u0c30\u0c3f\u0c2f\u0c41
\ntan C = \\(\\frac{\\mathrm{AB}}{\\mathrm{BC}}=\\frac{5}{12}\\)
\n\\(\\frac{5}{12}=\\frac{5 \\mathrm{k}}{24}\\) \u21d2 k= 2
\nAB = 10 \u0c38\u0c46\u0c02.\u0c2e\u0c40.
\n\u0c15\u0c30\u0c4d\u0c23\u0c02 AC = \\(\\sqrt{\\mathrm{AB}^{2}+\\mathrm{BC}^{2}}=\\sqrt{10^{2}+24^{2}}\\)
\n= \\(\\sqrt{100+576}=\\sqrt{676}\\) = 26 \u0c38\u0c46\u0c02.\u0c2e\u0c40.
\n(\u0c32\u0c47\u0c26\u0c3e)
\ntan C = \\(\\frac{5}{12}\\) (5, 12, 13 \u0c32\u0c41 \u0c2a\u0c48\u0c25\u0c3e\u0c17\u0c30\u0c3f\u0c2f\u0c28\u0c4d \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c3e\u0c32\u0c41),
\nBC = 24 \u0c38\u0c46\u0c02.\u0c2e\u0c40. (10, 24, 26 \u0c32\u0c41 \u0c15\u0c42\u0c21\u0c3e \u0c2a\u0c48\u0c25\u0c3e\u0c17\u0c30\u0c3f\u0c2f\u0c28\u0c4d \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c3e\u0c32\u0c41)
\n\u2234 \u0c15\u0c30\u0c4d\u0c23\u0c02 AC = 26 \u0c38\u0c46\u0c02.\u0c2e\u0c40.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 117.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c28\u0c3f \u0c1c\u0c24\u0c2a\u0c30\u0c1a\u0c21\u0c02\u0c32\u0c4b \u0c38\u0c30\u0c48\u0c28 \u0c38\u0c2e\u0c3e\u0c27\u0c3e\u0c28\u0c2e\u0c41\u0c28\u0c41 \u0c0e\u0c28\u0c4d\u0c28\u0c41\u0c15\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\"AP
\n(A) i-a, ii-b, iii-c
\n(B) i-b, ii-c, iii-a
\n(C) i-c, ii-b, iii-a
\n(D) i-c, ii-a, iii-b
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41.
\n(C) i-c, ii-b, iii-a<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 118.
\n\u0c2a\u0c4d\u0c30\u0c35\u0c1a\u0c28\u0c02 P: sin \u03b1 = cos \u03b2 \u0c05\u0c2f\u0c3f\u0c28 \u03b1 + \u03b2 = 90\u00b0 (\u03b1, \u03b2 \u0c32\u0c41 \u0c05\u0c32\u0c4d\u0c2a\u0c15\u0c4b\u0c23\u0c3e\u0c32\u0c41)
\n\u0c2a\u0c4d\u0c30\u0c35\u0c1a\u0c28\u0c02 – Q : sec \u03b8 = m, tan \u03b8 = n \u0c05\u0c2f\u0c3f\u0c28 m2<\/sup> – 12 = 1 (0\u00b0 < \u03b8 < 90\u00b0)
\n(A) P \u0c38\u0c24\u0c4d\u0c2f\u0c02, Q \u0c05\u0c38\u0c24\u0c4d\u0c2f\u0c02
\n(B) P \u0c38\u0c24\u0c4d\u0c2f\u0c02, Q \u0c38\u0c24\u0c4d\u0c2f\u0c02
\n(C) P \u0c05\u0c38\u0c24\u0c4d\u0c2f\u0c02, Q \u0c38\u0c24\u0c4d\u0c2f\u0c02
\n(D) P \u0c05\u0c38\u0c24\u0c4d\u0c2f\u0c02 , Q \u0c05\u0c38\u0c24\u0c4d\u0c2f\u0c02 .
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41.
\n(B) P \u0c38\u0c24\u0c4d\u0c2f\u0c02, Q \u0c38\u0c24\u0c4d\u0c2f\u0c02<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 119.
\nsin 60\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35
\n(A) cos 30\u00b0
\n(B) \\(\\frac{\\sqrt{3}}{2}\\)
\n(C) A \u0c2e\u0c30\u0c3f\u0c2f\u0c41 B
\n(D) \\(\\frac{1}{2}\\)
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41.
\n(C) A \u0c2e\u0c30\u0c3f\u0c2f\u0c41 B<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 120.
\nA, B \u0c32\u0c41 \u0c32\u0c18\u0c41\u0c15\u0c4b\u0c23\u0c3e\u0c32\u0c48\u0c28 sin (A – B) = \\(\\frac{1}{2}\\) sin A = \\(\\frac{1}{2}\\) \u0c05\u0c2f\u0c3f\u0c28 B \u0c28\u0c3f \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin (A – B) = \\(\\frac{1}{2}\\) \u21d2 A – B = 30\u00b0
\nsin A = \\(\\frac{1}{2}\\) \u21d2 A = 30\u00b0
\n\u2234 30\u00b0 – B = 30\u00b0 \u21d2 B = 0\u00b0
\n(\u0c32\u0c47\u0c26\u0c3e)
\nsin (A – B) = sin A = \\(\\frac{1}{2}\\)
\n\u2234 A – B = A \u21d2 – B = A – A \u21d2 B = 0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 121.
\n\u2206 ABC \u0c32\u0c4b a = 3; b = 4; c=’5 \u0c05\u0c2f\u0c3f\u0c28 cos A \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\"AP
\ncos A = \\(\\frac{\\mathrm{AC}}{\\mathrm{AB}}\\) = \\(\\frac{4}{5}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 122.
\nsec A . \\(\\sqrt{1-\\sin ^{2} A}\\) = 1 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 123.
\nsec2<\/sup>35\u00b0 – co2<\/sup>55\u00b0 = 1 \u0c05\u0c28\u0c3f \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\nsec2<\/sup>35\u00b0 – cot2<\/sup> (90 \u2013 35) (\u2235 cot (90 – \u03b8) = tan \u03b8)
\nsec2<\/sup>35\u00b0 – tan2<\/sup>35\u00b0 = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 124.
\n\\(\\frac{\\sin 18^{\\circ}}{\\cos 72^{\\circ}}\\) \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c38\u0c3e\u0c27\u0c28.
\n\\(\\frac{\\sin 18^{\\circ}}{\\cos 72^{\\circ}}\\) = \\(\\frac{\\sin (90-72)}{\\cos 72^{\\circ}}\\) = \\(\\frac{\\cos 72^{\\circ}}{\\cos 72^{\\circ}}\\) = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 125.
\ncos 1\u00b0 \u2219 cos 2\u00b0 \u2219 cos 3\u00b0…….. cos 90\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c38\u0c3e\u0c27\u0c28.
\ncos 1\u00b0 – cos 2\u00b0 – cos 3\u00b0 …….. cos 90\u00b0 = 0 (\u2235 cos 90\u00b0 = 0)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 126.
\ncot \u03b8 = p, sec \u03b8 = q \u0c05\u0c2f\u0c3f\u0c28 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b \u0c0f\u0c26\u0c3f \u0c38\u0c24\u0c4d\u0c2f\u0c02 ?
\n(A) p2<\/sup> – q2<\/sup> = 1
\n(B) q2<\/sup> – p2<\/sup> = 1
\n(C) q2<\/sup> – \\(\\frac{1}{\\mathrm{p}^{2}}\\) = 1
\n(D) \\(\\frac{1}{\\mathrm{q}^{2}}\\) – p2<\/sup> = 1
\n\u0c38\u0c3e\u0c27\u0c28.
\nC.
\ncot \u03b8 = p \u21d2 tan \u03b8 = \\(\\frac{1}{\\mathrm{p}}\\)
\nsec2<\/sup> \u03b8 – tan2<\/sup> \u03b8 = q2<\/sup> – \\(\\left(\\frac{1}{p}\\right)^{2}\\) = q2<\/sup> – \\(\\frac{1}{\\mathrm{p}}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 127.
\n\u0c12\u0c15 \u0c32\u0c02\u0c2c\u0c15\u0c4b\u0c23 \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c \u0c2d\u0c41\u0c1c\u0c3e\u0c32\u0c41 \u0c2a\u0c42\u0c30\u0c4d\u0c23\u0c3e\u0c02\u0c15\u0c3e\u0c32\u0c41 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 tan \u03b8 = \\(\\frac{5}{12}\\), \u0c05\u0c2f\u0c3f\u0c24\u0c47 \u0c06 \u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c\u0c2e\u0c41 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c15\u0c28\u0c3f\u0c37\u0c4d\u0c20 \u0c35\u0c48\u0c36\u0c3e\u0c32\u0c4d\u0c2f\u0c2e\u0c41 \u0c0e\u0c02\u0c24 ?
\n\u0c38\u0c3e\u0c27\u0c28.
\n\"AP
\n\u0c24\u0c4d\u0c30\u0c3f\u0c2d\u0c41\u0c1c\u0c2e\u0c41 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c15\u0c28\u0c3f\u0c37\u0c4d\u0c1f \u0c35\u0c48\u0c36\u0c3e\u0c32\u0c4d\u0c2f\u0c2e\u0c41 = \\(\\frac{1}{2}\\)ab
\n= \\(\\frac{1}{2}\\) \u00d7 5 \u00d7 12 = 30 \u0c1a.\u0c38\u0c46\u0c02.\u0c2e\u0c40.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 128.
\n\\(\\frac{\\sqrt{1+\\tan ^{2} \\theta}}{\\sqrt{1+\\cot ^{2} \\theta}}\\) = tan \u03b8 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41
\n\u0c38\u0c3e\u0c27\u0c28.
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 129.
\nsin2<\/sup> 47\u00b0 + sin2<\/sup> 43\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin2<\/sup> 47\u00b0 + sin2<\/sup> 43\u00b0 .
\n= sin2<\/sup> 47\u00b0 + sin2<\/sup> (90 – 47)\u00b0
\n= sin2<\/sup> 47\u00b0 + cos2<\/sup> 47\u00b0 = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 130.
\ncos \\(\\left(\\frac{\\theta}{2}\\right)\\) = \\(\\frac{1}{\\sqrt{2}}\\) \u0c05\u0c2f\u0c3f\u0c28 cos \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c38\u0c3e\u0c27\u0c28.
\ncos \\(\\left(\\frac{\\theta}{2}\\right)\\) = \\(\\frac{1}{\\sqrt{2}}\\) \u21d2 \\(\\frac{\\theta}{2}\\) = 45\u00b0 \u21d2 \u03b8 = 90\u00b0
\n\u2234 cos \u03b8 = cos 90\u00b0 = 0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 131.
\nsin \u03b8 = \\(\\frac{\\sqrt{3}}{2}\\) \u0c05\u0c2f\u0c3f\u0c28 sin \\(\\left(\\frac{\\theta}{2}\\right)\\) \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin \u03b8 = \\(\\frac{\\sqrt{3}}{2}\\) \u21d2 \u03b8 = 60\u00b0 \u2234 \\(\\frac{\\theta}{2}\\) = 30\u00b0
\nsin \\(\\frac{\\theta}{2}\\) = sin 30\u00b0 = \\(\\frac{1}{2}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 132.
\n(1 + cot2<\/sup> 45\u00b0)2<\/sup> \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c38\u0c3e\u0c27\u0c28.
\n(1 + cot2<\/sup> 45\u00b0)2<\/sup> = (1 + 12<\/sup>)2<\/sup>
\n= (1 + 1)2<\/sup> = 22<\/sup> = 4<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 133.
\n\\(\\frac{1-\\tan ^{2} 30^{\\circ}}{1+\\tan ^{2} 30^{\\circ}}\\) \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c38\u0c3e\u0c27\u0c28.
\n\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 134.
\n\\(\\frac{{cosec}^{2} \\theta}{\\cot \\theta}\\) – cot \u03b8 = tan \u03b8 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 135.
\n\\(\\sqrt{\\frac{\\sec x+\\tan x}{\\sec x-\\tan x}}\\) =
\n(A) sec x + tan x
\n(B) sec x – tan x
\n(C) 2 tan x
\n(D) 2 sec x
\n\u0c38\u0c3e\u0c27\u0c28.
\n(A) sec x + tan x<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 136.
\n4 sin 30\u00b0 \u2219 sec 60\u00b0 = x tan 45\u00b0 \u0c05\u0c2f\u0c3f\u0c28 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24?
\n\u0c38\u0c3e\u0c27\u0c28.
\n4 sin 30\u00b0 \u2219 sec 60\u00b0 = x tan 45\u00b0
\n\u21d2 4 \u00d7 \\(\\left(\\frac{1}{2}\\right)\\)(2) = x(1) \u21d2 4 = x
\n\u2234 x = 4<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 137.
\nsin 60\u00b0 cos 30\u00b0 + cos 60\u00b0. sin 30\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24?
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin 60\u00b0 . cos 30\u00b0 + cos 60\u00b0. sin 30\u00b0
\n= \\(\\left(\\frac{\\sqrt{3}}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right)+\\left(\\frac{1}{2}\\right)\\left(\\frac{1}{2}\\right)\\)
\n= \\(\\frac{3}{4}+\\frac{1}{4}=\\frac{4}{4}\\) = 1
\n(\u0c32\u0c47\u0c26\u0c3e)
\nsin 60\u00b0 \u00b7 cos 30\u00b0 + cos 60\u00b0. sin 30\u00b0
\n= sin (60\u00b0 + 30\u00b0) = sin 90\u00b0 = 1<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 138.
\nsin (A + B) = \\(\\frac{\\sqrt{3}}{2}\\); cos B = \\(\\frac{\\sqrt{3}}{2}\\) \u0c05\u0c2f\u0c3f\u0c28 A \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin (A + B) = \\(\\frac{\\sqrt{3}}{2}\\), cos B = \\(\\frac{\\sqrt{3}}{2}\\)
\nA + B = 60\u00b0, B = 30\u00b0
\nA + 30\u00b0 = 60\u00b0 \u21d2 A = 30\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 139.
\n\\(\\frac{\\tan 45^{\\circ}}{{cosec} 30^{\\circ}}+\\frac{\\sec 60^{\\circ}}{\\cot 45^{\\circ}}\\) =
\n(A) 2\\(\\frac{1}{2}\\)
\n(B) \\(\\frac{5}{2}\\)
\n(C) 2
\n(D) A \u0c2e\u0c30\u0c3f\u0c2f\u0c41 B
\n\u0c38\u0c3e\u0c27\u0c28.
\n(D) A \u0c2e\u0c30\u0c3f\u0c2f\u0c41 B
\n\\(\\frac{\\tan 45^{\\circ}}{{cosec} 30^{\\circ}}+\\frac{\\sec 60^{\\circ}}{\\cot 45^{\\circ}}\\) = \\(\\frac{1}{2}+\\frac{2}{1}\\) = 2\\(\\frac{1}{2}\\) = \\(\\frac{5}{2}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 140.
\ncos (A – B) = \\(\\frac{1}{2}\\); sin B = \\(\\frac{1}{\\sqrt{2}}\\) \u0c05\u0c2f\u0c3f\u0c28 A \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\nCOS (A – B) = \\(\\frac{1}{2}\\), sin B = \\(\\frac{1}{\\sqrt{2}}\\)
\nA – B = 60\u00b0,
\nB = 45\u00b0
\n\u2234 A = 60\u00b0 + B = 60\u00b0 + 45\u00b0 = 105\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 141.
\ncos 75\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b \u0c26\u0c47\u0c28\u0c3f\u0c15\u0c3f \u0c38\u0c2e\u0c3e\u0c28\u0c2e\u0c41 ?
\n(A) sin 15\u00b0
\n(B) – sin 15\u00b0
\n(C) cos 15\u00b0
\n(D) \\(\\frac{\\sqrt{3}}{2}\\)
\n\u0c38\u0c3e\u0c27\u0c28.
\n(A) sin 15\u00b0
\ncos 75\u00b0 = cos (90 – 15)\u00b0 = sin 15.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 142.
\n(1 + tan2<\/sup> A) (1 – sin2<\/sup> A) = 1 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\n(1 + tan2<\/sup> A) (1 – sin2<\/sup> A)
\n= sec2<\/sup> A – cos2<\/sup> A
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 143.
\ntan 30\u00b0 =
\n(A) \u221a3
\n(B) cot 60\u00b0
\n(C) \\($\\frac{1}{\\sqrt{3}}$\\)
\n(D) B \u0c2e\u0c30\u0c3f\u0c2f\u0c41 C .
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41.
\n(D) B \u0c2e\u0c30\u0c3f\u0c2f\u0c41 C .<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 144.
\nsin2<\/sup>45\u00b0 + cos2<\/sup>45\u00b0 + tan2<\/sup>45\u00b0 = 2 \u0c05\u0c28\u0c3f \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c3f\u0c02\u0c1a\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin2<\/sup>45\u00b0 + cos2<\/sup>45\u00b0 + tan2<\/sup>45\u00b0
\n= 1 + 1 = 2 [\u2235 sin2<\/sup>\u03b8 + cos2<\/sup> \u03b8 = 1]<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 145.
\n\\(\\frac{\\sin ^{4} A-\\cos ^{4} A}{\\sin ^{2} A-\\cos ^{2} A}\\) = 1 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 146.
\nsin \u03b8 = \\(\\frac{1}{2}\\) \u0c05\u0c2f\u0c3f\u0c28 cos \\(\\frac{3 \\theta}{2}\\) \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin \u03b8 = \\(\\frac{1}{2}\\) \u21d2 \u03b8 = 30\u00b0
\ncos \\(\\left(\\frac{3 \\theta}{2}\\right)\\) = cos\\(\\left(\\frac{3(30)}{2}\\right)\\)
\n= cos 45\u00b0 = \\(\\frac{1}{\\sqrt{2}}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 147.
\ncos (90 – \u03b8) =
\n(A) cos \u03b8
\n(B) sin \u03b8
\n(C) – sin \u03b8
\n(D) sec \u03b8
\n\u0c38\u0c3e\u0c27\u0c28.
\n(B) sin \u03b8<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 148.
\n\\(\\frac{\\tan \\theta \\cdot \\sqrt{1-\\sin ^{2} \\theta}}{\\sqrt{1-\\cos ^{2} \\theta}}\\) = 1 \u0c05\u0c28\u0c3f \u0c28\u0c3f\u0c30\u0c41\u0c2a\u0c3f\u0c02\u0c1a\u0c41\u0c2e\u0c41
\n\u0c38\u0c3e\u0c27\u0c28.
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 149.
\nsec \u03b8 + tan \u03b8 = \\(\\frac{1}{5}\\) \u0c05\u0c2f\u0c3f\u0c28 sin \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4b\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 150.
\n(sec 45\u00b0 + tan 45\u00b0) (sec 45\u00b0 – tan 45\u00b0) \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24?
\n\u0c38\u0c3e\u0c27\u0c28.
\n(sec 45\u00b0 + tan 45\u00b0) (sec 45\u00b0 – tan 45\u00b0)
\n= (sec2<\/sup> 45\u00b0 – tan2<\/sup> 45\u00b0)
\n= 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 151.
\nsin 30\u00b0 + cos 60\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24?
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin 30\u00b0 + cos 60\u00b0 = \\(\\frac{1}{2}\\) + \\(\\frac{1}{2}\\) = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 152.
\n\\(\\sqrt{{cosec}^{2} \\theta-\\cot ^{2} \\theta}\\) \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24?
\n\u0c38\u0c3e\u0c27\u0c28.
\n\\(\\sqrt{{cosec}^{2} \\theta-\\cot ^{2} \\theta}\\) = \u221a1 = \u00b1 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 153.
\ntan A \u0c28\u0c41 cos A \u0c32\u0c4b \u0c30\u0c3e\u0c2f\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\ntan A = \\(\\frac{\\sin A}{\\cos A}=\\frac{\\sqrt{1-\\cos ^{2} A}}{\\cos A}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 154.
\n\\(\\frac{1}{1-\\sin \\theta}+\\frac{1}{1+\\sin \\theta}\\) = 2 sec2<\/sup> \u03b8 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41
\n\u0c38\u0c3e\u0c27\u0c28.
\n\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 155.
\nsec \u03b8 + tan \u03b8 = 4 \u0c05\u0c2f\u0c3f\u0c28 cos \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 156.
\ncot2<\/sup> \u03b8 = 3 \u0c05\u0c2f\u0c3f\u0c28 cosec \u03b8 = i.
\n\u0c38\u0c3e\u0c27\u0c28.
\ncot2<\/sup> \u03b8 = 3
\ncosec2<\/sup> \u03b8 – cot2<\/sup> \u03b8 = 1
\n\u21d2 cosec2<\/sup>\u03b8 – 3 = 1
\n\u21d2 cosec2<\/sup> \u03b8 = 1 + 3 = 4
\n\u21d2 cosec \u03b8 = 4 = 2
\n(\u0c32\u0c47\u0c26\u0c3e)
\ncot2<\/sup> \u03b8 = 3 \u21d2 cot \u03b8 = \u221a3 \u21d2 \u03b8 = 30\u00b0
\n\u2234 cosec \u03b8 = cosec 30\u00b0 = 2<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 157.
\ntan 0\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c30\u0c3e\u0c2f\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\ntan 0\u00b0 = 0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 158.
\ncos \u03b8 = – cos \u03b8 \u0c05\u0c2f\u0c3f\u0c28 \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24?
\n\u0c38\u0c3e\u0c27\u0c28.
\ncos \u03b8 = – cos \u03b8
\n\u21d2 cos \u03b8 + cos \u03b8 = 0
\n\u21d2 cos \u03b8 = \\(\\frac{0}{2}\\) = 0
\n\u2234 \u03b8 = 90\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 159.
\nsin \u03b8 = \\(\\frac{a}{b}\\); cos \u03b8 = \\(\\frac{c}{d}\\) \u0c05\u0c2f\u0c3f\u0c28 cot \u03b8 =
\n(A) \\(\\frac{\\mathrm{ab}}{\\mathrm{cd}}\\)
\n(B) \\(\\frac{\\mathrm{bc}}{\\mathrm{ad}}\\)
\n(C) \\(\\frac{c a}{b d}\\)
\n(D) \\(\\frac{\\mathrm{ad}}{\\mathrm{bc}}\\)
\n\u0c38\u0c3e\u0c27\u0c28.
\n(B) \\(\\frac{\\mathrm{bc}}{\\mathrm{ad}}\\)<\/p>\n

\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 160.
\nA \u0c05\u0c32\u0c4d\u0c2a \u0c15\u0c4b\u0c23\u0c66 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 tan A = \\(\\frac{1}{\\sqrt{3}}\\) \u0c05\u0c2f\u0c3f\u0c28 sin A \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c38\u0c3e\u0c27\u0c28.
\ntan A = \\(\\frac{1}{\\sqrt{3}}\\) \u21d2 A = 30\u00b0
\n\u2234 sin A = sin 30\u00b0 = \\(\\frac{1}{2}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 161.
\na sin \u03b8 = x; b cos \u03b8 = y \u0c05\u0c2f\u0c3f\u0c28 \\(\\left(\\frac{x}{a}\\right)^{2}+\\left(\\frac{y}{b}\\right)^{2}\\) =
\n(A) – 1
\n(B) 0
\n(C) 2
\n(D) 1
\n\u0c38\u0c3e\u0c27\u0c28.
\n(D) 1
\na sin \u03b8 = x, b cos \u03b8 = y
\nsin \u03b8 = \\(\\frac{x}{a}\\) cos \u03b8 = \\(\\frac{y}{b}\\)
\n\u2234 \\(\\left(\\frac{x}{a}\\right)^{2}+\\left(\\frac{y}{b}\\right)^{2}\\) = sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 162.
\nsin A = \\(\\frac{1}{\\sqrt{2}}\\) \u0c05\u0c2f\u0c3f\u0c28 tan A \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24>
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin A = \\(\\frac{1}{\\sqrt{2}}\\) \u21d2 A = 45\u00b0
\n\u2234 tan A = tan 45\u00b0 = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 163.
\ncos (x – y) =
\n(A) cos x sin x + cos y sin y
\n(B) cos x sin y + cos y sin x
\n(C) sin x cos y + cos x sin y
\n(D) cos x cos y + sin x sin y
\n\u0c38\u0c3e\u0c27\u0c28.
\n(D) cos x cos y + sin x sin y
\ncos (x – y) = cos x cos y + sin x sin y<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 164.
\nsec \u03b8 = cosec \u03b8 \u0c05\u0c2f\u0c3f\u0c28 \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 =
\n(A) 90\u00b0
\n(B) 45\u00b0
\n(C) 30\u00b0
\n(D) 60\u00b0
\n\u0c38\u0c3e\u0c27\u0c28.
\n(B) 45\u00b0<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 165.
\n\\(\\frac{\\sqrt{{cosec}^{2} \\theta-1}}{{cosec} \\theta}\\) =
\n(A) 1 + sec \u03b8
\n(B) cosec \u03b8 + cot \u03b8
\n(C) cos \u03b8
\n(D) tan \u03b8
\n\u0c38\u0c3e\u0c27\u0c28.
\n(C) cos \u03b8<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 166.
\n\\(\\frac{\\sin \\theta}{\\sqrt{1-\\sin ^{2} \\theta}}\\) = tan \u03b8 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\\(\\frac{\\sin \\theta}{\\sqrt{1-\\sin ^{2} \\theta}}\\) = \\(\\frac{\\sin \\theta}{\\cos \\theta}\\) = tan \u03b8<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 167.
\ncos A = sin B \u0c05\u0c2f\u0c3f\u0c28 A + B = 90\u00b0 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\ncos A = sin B
\nsin (90 – A) = sin B \u21d2 90 – A = B
\n\u2234 90\u00b0 = A + B<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 168.
\nx = 2 cosec \u03b8; y = 2 cot \u03b8 \u0c05\u0c2f\u0c3f\u0c28 x2<\/sup> – y2<\/sup> \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24?
\n\u0c38\u0c3e\u0c27\u0c28.
\nx = 2 cosec \u03b8, y = 2 cot \u03b8
\nx2<\/sup> – y2<\/sup> = (2 cosec \u03b8)2<\/sup> \u2013 (2 cot \u03b8)2<\/sup>
\n= 4 cosec2<\/sup> \u03b8 – 4 cot2<\/sup> \u03b8
\n= 4 (cosec2<\/sup> \u03b8 – cot2<\/sup> \u03b8)
\n= 4 (1) = 4<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 169.
\ncos (A + B) = 0, cos B = \\(\\frac{\\sqrt{3}}{2}\\) \u0c05\u0c2f\u0c3f\u0c28 A \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\ncos (A + B) = 0, cos B = \\(\\frac{\\sqrt{3}}{2}\\)
\n\u2234 A + B = 90\u00b0 B = 30\u00b0
\n\u2234 A + 30\u00b0 = 90\u00b0 \u21d2 A = 90\u00b0 \u2013 30 = 60\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 170.
\n\\(\\frac{\\sqrt{1-\\cos ^{2} \\theta}}{\\cos \\theta}\\) =
\n(A) \\(\\frac{\\tan \\theta}{1+\\tan \\theta}\\)
\n(B) \\(\\frac{1}{\\tan \\theta}\\)
\n(C) cot \u03b8
\n(D) sin \u03b8
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41.
\n(D) sin \u03b8<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 171.
\n\\(\\frac{\\sqrt{1-\\cos ^{2} \\theta}}{\\cos \\theta}\\) =
\n(A) tan \u03b8
\n(B) cos \u03b8
\n(C) sec \u03b8
\n(D) cot \u03b8
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41.
\n(A) tan \u03b8<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 172.
\n\\(\\frac{1}{\\sqrt{1+\\tan ^{2} \\theta}}\\) =
\n(A) sin \u03b8
\n(B) cos \u03b8
\n(C) sec \u03b8
\n(D) cosec \u03b8
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41.
\n(B) cos \u03b8<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 173.
\ncos \u03b8 . tan \u03b8 =
\n(A) cos \u03b8
\n(B) cot \u03b8
\n(C) sin \u03b8
\n(D) cos2<\/sup> \u03b8<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 174.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c2a\u0c1f\u0c02 \u2206 ABC \u0c32\u0c4b \u2220C = 45\u00b0 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u2220B = 90\u00b0 \u0c05\u0c2f\u0c3f\u0c28 cos A \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24?
\n\"AP
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u2206 ABC \u0c32\u0c4b \u2220B = 90\u00b0 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 C = 45\u00b0 \u0c15\u0c3e\u0c35\u0c41\u0c28,
\n\u2234 \u2220A = 45\u00b0; cos A = cos 45\u00b0 = \\(\\frac{1}{\\sqrt{2}}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 175.
\nsec2<\/sup> 33\u00b0 – cot2<\/sup> 57\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\nsec2<\/sup> 33\u00b0 – cot2<\/sup> 57\u00b0
\n= sec2<\/sup> (90 – 57) – cot2<\/sup> 57\u00b0 (\u2235 sec (90 – \u03b8) = cosec \u03b8)
\n= cosec2<\/sup> 57\u00b0 – cot2<\/sup> 57\u00b0
\n= 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 176.
\n\u221a3 cot 2\u03b8 = 1 \u0c05\u0c2f\u0c3f\u0c28 \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u221a3 cot 2\u03b8 = 1 \u21d2 cot 2\u03b8 = \\(\\frac{1}{\\sqrt{3}}\\)
\n\u2234 2\u03b8 = 60\u00b0 (\u2235 cot 60\u00b0 = \\(\\frac{1}{\\sqrt{3}}\\))
\n\u2234 \u03b8 = \\(\\frac{60^{\\circ}}{2}\\) = 30\u00b0<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 177.
\nsin (log10<\/sub>1) \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin (log10<\/sub>1)\u00b0 = sin 0\u00b0 = 0 (\u2235 loga<\/sub> 1 = 0)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 178.
\ncos \u03b8 = 3 \u0c05\u0c2f\u0c3f\u0c28 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b \u0c0f\u0c26\u0c3f \u0c38\u0c24\u0c4d\u0c2f\u0c02 ?
\n(A) \u03b8 = 60\u00b0
\n(B) cos\\(\\left(\\frac{\\theta}{2}\\right)=\\frac{\\sqrt{3}}{2}\\)
\n(C) sin\\(\\left(\\frac{3 \\theta}{4}\\right)^{\\circ}=\\frac{1}{\\sqrt{2}}\\)
\n(D) \u0c2a\u0c48\u0c35\u0c28\u0c4d\u0c28\u0c40
\n\u0c38\u0c3e\u0c27\u0c28.
\n(D) \u0c2a\u0c48\u0c35\u0c28\u0c4d\u0c28\u0c40<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 179.
\ncos \u03b8 \u0c28\u0c41 sin \u03b8 \u0c32\u0c4b \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\ncos \u03b8 = \\(\\sqrt{1-\\sin ^{2} \\theta}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 180.
\ncos 0\u00b0 + sin 90\u00b0 + \u221a3 cosec 60\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\ncos 0\u00b0 + sin 90\u00b0 + \u221a3 cosec 60\u00b0
\n\"AP
\n= 1 + 1 + 2 = 4<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 181.
\ncos 60\u00b0 \u2219 cos 30\u00b0 – sin60\u00b0 \u2219 sin30\u00b0\u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\ncos 60\u00b0 \u2219 cos 30\u00b0 – sin 60\u00b0 \u2219 sin 30\u00b0
\n= \\(\\frac{1}{2} \\times \\frac{\\sqrt{3}}{2}-\\frac{\\sqrt{3}}{2} \\times \\frac{1}{2}=\\frac{\\sqrt{3}}{4}-\\frac{\\sqrt{3}}{4}\\) = 0
\n(\u0c32\u0c47\u0c26\u0c3e)
\ncos 60\u00b0 \u2219 Cos 30\u00b0 – sin 60\u00b0 \u2219 sin 30
\n= cos (60 + 30)\u00b0
\n= cos 90\u00b0 = 0
\n(\u2235cos (A + B) = cos A cos B – sin A sin B)<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 182.
\ntan \u03b8 =
\n(A) \\(\\frac{1}{\\cot \\theta}\\)
\n(B) \\(\\sqrt{\\sec ^{2} \\theta+1}\\)
\n(C) \\(\\frac{\\sin \\theta}{\\cos \\theta}\\)
\n(D) \u0c2a\u0c48\u0c35\u0c28\u0c4d\u0c28\u0c40
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41.
\n(D) \u0c2a\u0c48\u0c35\u0c28\u0c4d\u0c28\u0c40<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 183.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c40\u0c24\u0c3f\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c3f \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c15\u0c41 \u0c38\u0c2e\u0c3e\u0c28\u0c2e\u0c2a\u0c41 \u0c38\u0c02\u0c35\u0c30\u0c4d\u0c17\u0c2e\u0c3e\u0c28 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c1c\u0c24 \u0c1a\u0c47\u0c2f\u0c02\u0c21\u0c3f.
\n\"AP
\n(A) i-b, ii-a, iii-c, iv-d.
\n(B) i-c, ii-d, iii-b, iv-a
\n(C) i-a, ii-c, iii-d, iv-b
\n(D) i-d, ii-c, iii-b, iv-a
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41.
\n(D) i-d, ii-c, iii-b, iv-a<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 184.
\n\u221a3 tan \u03b8 = 1 \u0c05\u0c2f\u0c3f\u0c28 6 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ? (\u03b8 \u0c05\u0c32\u0c4d\u0c2a\u0c15\u0c4b\u0c23\u0c2e\u0c41)
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u221a3 tan \u03b8 = 1 \u21d2 tan \u03b8 = \\(\\frac{1}{\\sqrt{3}}\\)
\n\u2234 \u03b8 = 30\u00b0 (\u2234 tan 30\u00b0 = \\(\\frac{1}{\\sqrt{3}}\\))<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 185.
\n\\(\\sqrt{{cosec}^{2} \\theta-\\sin ^{2} \\theta-\\cos ^{2} \\theta}\\) \u0c28\u0c41 \u0c12\u0c15\u0c47 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c40\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c3f\u0c32\u0c4b \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 186.
\n\\(\\sqrt{\\frac{1-\\cos ^{2} \\theta}{1+\\cot ^{2} \\theta}}\\) =
\n(A) sin \u03b8
\n(B) sin \u03b8
\n(C) sin \u03b8
\n(D) sin4<\/sup> \u03b8
\n\u0c38\u0c3e\u0c27\u0c28.
\n(C) sin \u03b8
\n\\(\\sqrt{\\frac{1-\\cos ^{2} \\theta}{1+\\cot ^{2} \\theta}}=\\sqrt{\\frac{\\sin ^{2} \\theta}{{cosec}^{2} \\theta}}\\)
\n= \\(\\sqrt{\\sin ^{2} \\theta \\cdot \\sin ^{2} \\theta}\\) = sin2<\/sup> \u03b8<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 187.
\ncosec 60\u00b0 \u00d7 cos 90\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c38\u0c3e\u0c27\u0c28.
\ncosec 60\u00b0 \u00d7 cos 90\u00b0 = \\(\\frac{2}{\\sqrt{3}}\\) \u00d7 0 = 0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 188.
\nsin 81\u00b0 =
\n(A) cos 9\u00b0
\n(B) cos 81\u00b0
\n(C) – cos 9o
\n(D) cosec 81\u00b0
\n\u0c38\u0c3e\u0c27\u0c28.
\n(A) cos 9\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 189.
\ntan \u03b8 = \u221a3 \u0c05\u0c2f\u0c3f\u0c24\u0c47 sec \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4b\u0c28\u0c41\u0c2e\u0c41
\n\u0c38\u0c3e\u0c27\u0c28.
\ntan \u03b8 = \u221a3 \u21d2 \u03b8 = 60\u00b0
\n\u2234 sec \u03b8 = sec 60\u00b0 = 2
\n(\u0c32\u0c47\u0c26\u0c3e)
\nsec2<\/sup> \u03b8 = 1 + tan2<\/sup> \u03b8
\n= 1 + (\u221a3)2<\/sup> = 1 + 3 = 4
\n\u2234 sec \u03b8 = \u221a4 = 2<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 190.
\ncot2<\/sup>\u03b8 – \\(\\frac{1}{\\sin ^{2} \\theta}\\) = – 1 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\ncot2<\/sup>\u03b8 – \\(\\frac{1}{\\sin ^{2} \\theta}\\) = cot2<\/sup>\u03b8 – cosec2<\/sup>\u03b8
\n= – (cosec2<\/sup> \u03b8 – cot2<\/sup> \u03b8) = – 1<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 191.
\n\\(\\frac{1}{\\sec \\theta-\\tan \\theta}\\) = …………….
\n(A) sec \u03b8 – tan \u03b8
\n(B) sec \u03b8 + tan \u03b8
\n(C) sec2<\/sup>\u03b8 – tan2<\/sup>\u03b8
\n(D) 1
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41.
\n(B) sec \u03b8 + tan \u03b8<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 192.
\nsin 29\u00b0 – cos 61\u00b0 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24?
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin 29\u00b0 – cos 61\u00b0
\n= sin (90 – 61)\u00b0 – cos 61\u00b0
\n= cos 61\u00b0 \u2013 cos 61\u00b0 = 0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 193.
\ntan 1\u00b0 \u2219 tan 2\u00b0 \u2219 tan 3\u00b0 ……….. tan 89\u00b0 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c38\u0c3e\u0c27\u0c28.
\ntan 1\u00b0 \u2219 tan 2\u00b0 \u2219 tan 3\u00b0……….. tan 89\u00b0
\n= tan (90 – 89)\u00b0 . tan (90 – 88)\u00b0 ………. tan 45\u00b0….. tan 88\u00b0 \u2219 tan 89\u00b0
\n= cot 89\u00b0 . cot 88\u00b0….. tan 45\u00b0….. tan 88\u00b0 \u2219 tan 89 = tan 45\u00b0 = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 194.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b \u0c38\u0c30\u0c48\u0c28 \u0c2a\u0c4d\u0c30\u0c35\u0c1a\u0c28\u0c3e\u0c32\u0c28\u0c41 \u0c0e\u0c28\u0c4d\u0c28\u0c41\u0c15\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n(i) 0 \u2264 \u03b8 \u2264 90 \u0c05\u0c2f\u0c3f\u0c28 0 \u2264 sin \u03b8 \u2264 1
\n(ii) tan A = cot B (A, B \u0c32\u0c41 \u0c05\u0c32\u0c4d\u0c2a\u0c15\u0c4b\u0c23\u0c3e\u0c32\u0c41) \u0c05\u0c2f\u0c3f\u0c24\u0c47 A + B = 90\u00b0
\n(iii) cosec \u03b8 = p, cot \u03b8 = q \u0c05\u0c2f\u0c3f\u0c28 p2<\/sup> + q2<\/sup> = 1
\n(iv) cos \u03b8 = x \u0c05\u0c2f\u0c3f\u0c28 sin \u03b8 = \\(\\sqrt{1+x^{2}}\\).
\n\u0c38\u0c3e\u0c27\u0c28.
\n(i) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 (ii) \u0c32\u0c41 \u0c38\u0c24\u0c4d\u0c2f\u0c2e\u0c41.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 195.
\n\u0c2a\u0c48 194 \u0c35 \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28\u0c32\u0c4b \u0c05\u0c38\u0c24\u0c4d\u0c2f \u0c2a\u0c4d\u0c30\u0c35\u0c1a\u0c28\u0c3e\u0c32\u0c28\u0c41 \u0c17\u0c41\u0c30\u0c4d\u0c24\u0c3f\u0c02\u0c1a\u0c3f, \u0c35\u0c3e\u0c28\u0c3f\u0c28\u0c3f \u0c38\u0c24\u0c4d\u0c2f \u0c2a\u0c4d\u0c30\u0c35\u0c1a\u0c28\u0c3e\u0c32\u0c41\u0c17\u0c3e \u0c2e\u0c3e\u0c30\u0c4d\u0c1a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\n(iii) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 (iv) \u0c32\u0c41 \u0c05\u0c38\u0c24\u0c4d\u0c2f \u0c2a\u0c4d\u0c30\u0c35\u0c1a\u0c28\u0c3e\u0c32\u0c41.
\n\u0c35\u0c3e\u0c28\u0c3f\u0c28\u0c3f \u0c38\u0c24\u0c4d\u0c2f \u0c2a\u0c4d\u0c30\u0c35\u0c1a\u0c28\u0c3e\u0c32\u0c41\u0c17\u0c3e \u0c2e\u0c3e\u0c30\u0c4d\u0c1a\u0c17\u0c3e,
\ncosec \u03b8 = p, cot \u03b8 = q \u0c05\u0c2f\u0c3f\u0c28 p2<\/sup> – q2<\/sup> = 1
\ncos \u03b8 = x \u0c05\u0c2f\u0c3f\u0c28 sin \u03b8 = \\(\\sqrt{1-x^{2}}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 196.
\ntan (15\u00b0 + B) = \u221a3 \u0c05\u0c2f\u0c3f\u0c24\u0c47 B \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24?
\n\u0c38\u0c3e\u0c27\u0c28.
\ntan (15\u00b0 + B) = \u221a3 \u21d2 15 + B = 60\u00b0
\n\u2234 B = 60\u00b0 – 15\u00b0 = 45\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 197.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b \u0c0f\u0c26\u0c3f sin e \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c15\u0c3e\u0c35\u0c1a\u0c4d\u0c1a\u0c41\u0c28\u0c41 ? .
\n(A) \u221a2
\n(B) \\(\\frac{1}{\\sqrt{2}}\\)
\n(C) \u221a3
\n(D) 2
\n\u0c38\u0c3e\u0c27\u0c28.
\nB
\n(\u2235 \u221a2 = 1.414 ….. > 1,
\n\u221a3 = 1.732 ….. > 1
\n2 > 1)<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 198.
\n\\(\\sqrt{(\\sec \\theta+1)(\\sec \\theta-1)}\\) = tan \u03b8 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\\(\\sqrt{(\\sec \\theta+1)(\\sec \\theta-1)}\\) = \\(\\sqrt{\\sec ^{2} \\theta-1}\\)
\n= tan \u03b8<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 199.
\ncosec \u03b8 + cot \u03b8 = 2 \u0c05\u0c2f\u0c3f\u0c28 cosec \u03b8 – cot \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c38\u0c3e\u0c27\u0c28.
\ncosec \u03b8 + cot \u03b8 = 2 \u0c05\u0c2f\u0c3f\u0c28
\ncosec \u03b8 – cot \u03b8 = \\(\\frac{1}{2}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 200.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b cos \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c15\u0c3e\u0c28\u0c3f\u0c26\u0c3f \u0c0f\u0c26\u0c3f ?
\n(A) \\(\\frac{3}{4}\\)
\n(B) \\(\\frac{1}{2}\\)
\n(C) \\(\\frac{4}{3}\\)
\n(D) 0
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41.
\n(C) \\(\\frac{4}{3}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 201.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b \u0c0f\u0c26\u0c3f \u0c38\u0c24\u0c4d\u0c2f\u0c02 ?
\n(A) sin2<\/sup>\u03b8 + cos2<\/sup>\u03b8 = 1
\n(B) sin2<\/sup> \u03b8 = 1 – cos2<\/sup> \u03b8
\n(C) cos2<\/sup> \u03b8 = 1 – sin2<\/sup> \u03b8
\n(D) \u0c2a\u0c48\u0c35\u0c28\u0c4d\u0c28\u0c40
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41.
\n(D) \u0c2a\u0c48\u0c35\u0c28\u0c4d\u0c28\u0c40<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 202.
\nsin A = \\(\\frac{1}{\\sqrt{2}}\\), tan B = 1 \u0c05\u0c2f\u0c3f\u0c28 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c0f\u0c26\u0c3f \u0c38\u0c24\u0c4d\u0c2f\u0c02 \u0c0f\u0c26\u0c3f \u0c38\u0c24\u0c4d\u0c2f\u0c02 ? (A, B \u0c32\u0c41 \u0c05\u0c32\u0c4d\u0c2a\u0c15\u0c4b\u0c23\u0c3e\u0c32\u0c41)
\n(i) A = B
\n(ii) A + B = 60\u00b0
\n(iii) sin (A + B) = 1
\n(iv) cos (A + B) = 1
\n(A) i \u0c2e\u0c30\u0c3f\u0c2f\u0c41 ii
\n(B) i \u0c2e\u0c30\u0c3f\u0c2f\u0c41 iii
\n(C) ii \u0c2e\u0c30\u0c3f\u0c2f\u0c41 iv
\n(D) ii \u0c2e\u0c30\u0c3f\u0c2f\u0c41 iii
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41.
\n(B) i \u0c2e\u0c30\u0c3f\u0c2f\u0c41 iii<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 203.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c2a\u0c1f\u0c02\u0c32\u0c4b AB \u0c2a\u0c4a\u0c21\u0c35\u0c41\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\"AP
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u2206ABC\u0c32\u0c4b \u2220B = 90\u00b0
\n\u2234 tan 30\u00b0 = \\(\\frac{\\mathrm{AB}}{\\mathrm{BC}}\\) = \\(\\frac{A B}{60}\\)
\n\\(\\frac{1}{\\sqrt{3}}=\\frac{\\mathrm{AB}}{60}\\) \u21d2 AB = \\(\\frac{60}{\\sqrt{3}}\\) \u0c2e\u0c40.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 204.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b \u0c0f\u0c26\u0c3f \u0c28\u0c3f\u0c30\u0c4d\u0c35\u0c1a\u0c3f\u0c02\u0c2a\u0c2c\u0c21\u0c26\u0c41”?
\n(A) sin 45\u00b0
\n(B) cot 0\u00b0
\n(C) tan 0\u00b0
\n(D) sec 45\u00b0
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41.
\n(B) cot 0\u00b0<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 205.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b \u0c0f\u0c26\u0c3f \u0c38\u0c24\u0c4d\u0c2f\u0c02 ? ”
\n(A) sec2<\/sup>\u03b8 – tan2<\/sup>\u03b8 = 1
\n(B) sec2<\/sup>\u03b8 + tan2<\/sup>\u03b8 = 1
\n(c) tan2<\/sup>\u03b8 – sec2<\/sup> \u03b8 = 1
\n(D) \u0c2a\u0c48\u0c35\u0c28\u0c4d\u0c28\u0c40
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41.
\n(A) sec2<\/sup>\u03b8 – tan2<\/sup>\u03b8 = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 206.
\nA + B = 90\u00b0 \u0c05\u0c2f\u0c3f\u0c28 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b. \u0c0f\u0c26\u0c3f \u0c38\u0c24\u0c4d\u0c2f\u0c02 ?
\n(A) cos A = sin B
\n(B) tan A = cot B
\n(C) cosec A = sec A
\n(D) \u0c2a\u0c48\u0c35\u0c28\u0c4d\u0c28\u0c40
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41.
\n(D) \u0c2a\u0c48\u0c35\u0c28\u0c4d\u0c28\u0c40<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 207.
\nsec \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35 tan \u03b8 \u0c32\u0c4b \u0c30\u0c3e\u0c2f\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\nsec2<\/sup>\u03b8 – tan2<\/sup> \u03b8 = 1
\n\u21d2 sec2<\/sup> 8 = 1 + tan2<\/sup>\u03b8
\n\u21d2 sec \u03b8 = \\(\\sqrt{1+\\tan ^{2} \\theta}\\).<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 208.
\n\u0c1c\u0c24\u0c2a\u0c30\u0c41\u0c1a\u0c41\u0c2e\u0c41.
\n\"AP
\n(A) a – iii, b-i, c – iv, d – ii
\n(B) a – iv, b – ii, c-i, d – iii
\n(C) a – iii, b-ii, c – iv, d-i
\n(D) a – iv, b – iii, c-ii, d – i
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41.
\n(B) a – iv, b – ii, c-i, d – iii<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 209.
\nA, B\u0c32\u0c41 \u0c05\u0c32\u0c4d\u0c2a\u0c15\u0c4b\u0c23\u0c3e\u0c32\u0c41 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u2220A \u2265 \u2220B \u0c05\u0c2f\u0c3f\u0c28 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b \u0c0f\u0c26\u0c3f \u0c05\u0c38\u0c24\u0c4d\u0c2f\u0c02 ?
\n(A) sin A \u2265 sin B
\n(B) cos A \u2264 cos B
\n(C) tan A \u2264 tan B
\n(D) sec A \u2265 sec B
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41.
\nC) tan A < tan B<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 210.
\ncosec \u03b8, cot \u03b8 \u0c32\u0c32\u0c4b \u0c38\u0c30\u0c4d\u0c35\u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c3e\u0c28\u0c4d\u0c28\u0c3f \u0c30\u0c3e\u0c2f\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\ncosec2<\/sup>\u03b8 – cot- \u03b8 = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 211.
\nsin 75\u00b0 + cos 65\u00b0 \u0c28\u0c41 0\u00b0 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 45\u00b0 \u0c32 \u0c2e\u0c27\u0c4d\u0c2f\u0c17\u0c32 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c4b \u0c30\u0c3e\u0c2f\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin 75\u00b0 + cos 65\u00b0
\n= sin (90 – 15)\u00b0 + cos (90 – 25)
\n= cos 15\u00b0 + sin 25\u00b0<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 212.
\nsin 73\u00b0 + tan 73\u00b0\u0c28\u0c41 cos, cot \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c41\u0c32 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c32\u0c4b \u0c30\u0c3e\u0c2f\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin 73\u00b0 + tan 73\u00b0
\n= sin (90 – 17)\u00b0 + tan (90 – 17)\u00b0
\n= cos 17\u00b0 + cot 17\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 213.
\ntan \u03b8 = p, sec \u03b8 = q \u0c05\u0c2f\u0c3f\u0c28 p, q \u0c32 \u0c2e\u0c27\u0c4d\u0c2f \u0c38\u0c02\u0c2c\u0c02\u0c27\u0c2e\u0c41 ………………. .
\n(A) p2<\/sup> = 1 + q2<\/sup>
\n(B) q2<\/sup> = 1 + p2<\/sup>
\n(C) p2<\/sup> + q2<\/sup> = 1
\n(D) p2<\/sup>= 2q2<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n(B) q2<\/sup> = 1 + p2<\/sup>
\nsec2<\/sup> \u03b8 = 1 + tan2<\/sup>\u03b8
\n\u2234 q2<\/sup> = 1 + p2<\/sup><\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 214.
\ncosec \u03b8 = m, cot \u03b8 = n \u0c05\u0c2f\u0c3f\u0c28 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b \u0c0f\u0c26\u0c3f \u0c05\u0c38\u0c24\u0c4d\u0c2f\u0c02?
\n(A) m2<\/sup> – n2<\/sup> = 1
\n(B) sec \u03b8 = \\(\\frac{m}{n}\\)
\n(C) cos \u03b8 = \\(\\frac{n}{m}\\)
\n(D) m2<\/sup> + n2<\/sup> = 1
\n\u0c38\u0c3e\u0c27\u0c28.
\n(A) m2<\/sup> – n2<\/sup> = 1
\ncosec2<\/sup> \u03b8 – cot2<\/sup> \u03b8 = 1
\nm2<\/sup> – n2<\/sup> = 1<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 215.
\nsin \u03b8 = x \u0c05\u0c2f\u0c3f\u0c28 cos \u03b8 \u0c28\u0c41 x \u0c32\u0c4b \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\ncos \u03b8 = \\(\\sqrt{1-\\sin ^{2} \\theta}\\) = \\(\\sqrt{1-x^{2}}\\).
\n(\u2235 sin \u03b8 = x)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 216.
\nsin 0\u00b0, sin 60\u00b0, sin 30\u00b0, sin \u03b8, sin 90\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c2e\u0c27\u0c4d\u0c2f\u0c17\u0c24\u0c02 \\(\\frac{1}{\\sqrt{2}}\\) \u0c05\u0c2f\u0c4d\u0c2f\u0c47\u0c1f\u0c1f\u0c4d\u0c32\u0c41 \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\nsin 0\u00b0 = 0, sin 60\u00b0 = \\(\\frac{\\sqrt{3}}{2}\\), sin 30\u00b0 = \\(\\frac{1}{2}\\),
\nsin 90\u00b0 = 1, sin \u03b8 = ?
\nsin 0\u00b0, sin 60\u00b0, sin 30\u00b0, sin \u03b8, sin 90\u00b0 \u0c32 \u0c2e\u0c27\u0c4d\u0c2f\u0c17\u0c24\u0c2e\u0c41 = \\(\\frac{1}{\\sqrt{2}}\\)
\n\u2234 sin \u03b8 = \\(\\frac{1}{\\sqrt{2}}\\) \u21d2 \u03b8 = 45\u00b0
\n[\u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c28\u0c41 \u0c06\u0c30\u0c4b\u0c39\u0c23\u0c3e\u0c15\u0c4d\u0c30\u0c2e\u0c02\u0c32\u0c4b \u0c30\u0c3e\u0c2f\u0c17\u0c3e
\n0, \\(\\frac{1}{2}\\), sin \u03b8, \\(\\frac{\\sqrt{3}}{2}\\), 1.
\nsin \u03b8, \\(\\frac{1}{2}\\) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \\(\\frac{\\sqrt{3}}{2}\\) \u0c32 \u0c2e\u0c27\u0c4d\u0c2f \u0c09\u0c28\u0c4d\u0c28\u0c2a\u0c4d\u0c2a\u0c41\u0c21\u0c47 \u0c2e\u0c27\u0c4d\u0c2f\u0c17\u0c24\u0c02 \\(\\frac{1}{\\sqrt{2}}\\) \u0c05\u0c2f\u0c4d\u0c2f\u0c47 \u0c05\u0c35\u0c15\u0c3e\u0c36\u0c02 \u0c15\u0c32\u0c26\u0c41.)<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 217.
\nsin 09\u00b0, sec 60\u00b0, cos 0\u00b0, tan 45\u00b0 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 tan2<\/sup> \u03b8 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c38\u0c17\u0c1f\u0c41 \\(\\frac{7}{5}\\) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \\(\\frac{1}{2}\\) \u0c05\u0c32\u0c4d\u0c2a\u0c15\u0c4b\u0c23\u0c2e\u0c41 \u0c05\u0c2f\u0c3f\u0c28 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c38\u0c17\u0c1f\u0c41 = \\(\\frac{\\sin 0^{\\circ}+\\sec 60^{\\circ}+\\cos 0^{\\circ}+\\tan 45^{\\circ}+\\tan ^{2} \\theta}{5}\\)
\n\"AP
\n\u2234 4 + tan2<\/sup> \u03b8 = 7 \u21d2 tan2<\/sup> \u03b8 = 7 – 4 = 3
\n\u2234 tan \u03b8 = \u221a3 \u21d2 \u03b8 = 60\u00b0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 218.
\nsec \u03b8 + tan \u03b8 = x, sec \u03b8 – tan \u03b8 = y \u0c05\u0c2f\u0c3f\u0c28 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b \u0c0f\u0c26\u0c3f \u0c38\u0c24\u0c4d\u0c2f\u0c02 ?
\n(i) xy = 1
\n(ii) \\(\\frac{x}{y}\\) = 1
\n(iii) x2<\/sup> + y2<\/sup> = 1
\n(iv) x2<\/sup> – y2<\/sup> = 1
\n\u0c38\u0c3e\u0c27\u0c28.
\n(i) \u0c38\u0c24\u0c4d\u0c2f\u0c02
\n(sec \u03b8 + tan \u03b8) (sec \u03b8 – tan \u03b8) = xy
\nsec2<\/sup>\u03b8 – tan2<\/sup> \u03b8 = 1
\n(\u2234 (a + b) (a – b) = a2<\/sup> – b2<\/sup>)
\n\u2234 x \u2219 y = 1.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28219.
\n\u2220A \u0c15\u0c41 \u0c0e\u0c26\u0c41\u0c1f\u0c3f \u0c2d\u0c41\u0c1c\u0c02 \u0c0f\u0c26\u0c3f ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\u2220A \u0c15\u0c41 \u0c0e\u0c26\u0c41\u0c1f\u0c3f \u0c2d\u0c41\u0c1c\u0c02 = BC<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28220.
\n\u2220C \u0c15\u0c41 \u0c06\u0c38\u0c28\u0c4d\u0c28 \u0c2d\u0c41\u0c1c\u0c2e\u0c41\u0c28\u0c41 \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\u2220C \u0c15\u0c41 \u0c06\u0c38\u0c28\u0c4d\u0c28 \u0c2d\u0c41\u0c1c\u0c2e\u0c41 = BC<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28221.
\n\u2220A + \u2220C \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\u2220A + \u2220C = 90\u00b0 [\u2235 \u2220B = 90\u00b0, \u2220A + \u2220B + \u2220C = 180\u00b0]<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28222.
\n\\(\\frac{\\mathbf{B C}}{\\mathbf{A B}}\\) \u0c28\u0c3f \u0c38\u0c42\u0c1a\u0c3f\u0c02\u0c1a\u0c41 \u0c0f\u0c26\u0c47\u0c28\u0c3f \u0c12\u0c15 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c40\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c3f\u0c28\u0c3f \u0c24\u0c46\u0c32\u0c4d\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\ntan A (\u0c32\u0c47\u0c26\u0c3e) cot C<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28223.
\nsin A = cos B \u0c05\u0c28\u0c3f \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nA + B = 90\u00b0 \u21d2 A = 90 – B
\n\u2234 sin A = sin (90 – B) = cos B [\u2235 sin (90 – \u03b8) = cos \u03b8]
\n\u2234 sin A = cos B<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28224.
\n\\(\\frac{\\mathbf{B C}}{\\mathbf{A C}}\\) \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c3f\u0c15\u0c3f \u0c2a\u0c4d\u0c30\u0c3e\u0c24\u0c3f\u0c28\u0c3f\u0c25\u0c4d\u0c2f\u0c02 \u0c35\u0c39\u0c3f\u0c02\u0c1a\u0c41 \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c40\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c3f
\nA) tan A
\nB) sin A
\nC) cos C
\nD) B \u0c2e\u0c30\u0c3f\u0c2f\u0c41 C
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nD) B \u0c2e\u0c30\u0c3f\u0c2f\u0c41 C<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 225.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c32\u0c4b \u0c0f\u0c26\u0c3f \u0c38\u0c24\u0c4d\u0c2f\u0c02 ?
\nA) AC > BC
\nB) AC > AB
\nC) A + C = 90\u00b0
\nD) \u0c2a\u0c48\u0c35\u0c28\u0c4d\u0c28\u0c40
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nD) \u0c2a\u0c48\u0c35\u0c28\u0c4d\u0c28\u0c40<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28226.
\nsec 16. cosec 74\u00b0 – cot 74\u00b0 – tan 16\u00b0 \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c02\u0c21\u0c3f.
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n0 (\u0c38\u0c41\u0c28\u0c4d\u0c28)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28227.
\nx = 2019\u00b0 \u0c05\u0c2f\u0c3f\u0c24\u0c47 sin2<\/sup> x + cos2<\/sup> x \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\nx = 2019\u00b0, \u0c05\u0c2f\u0c3f\u0c24\u0c47 sin2<\/sup> x + cos2<\/sup> x = sin2<\/sup>2019\u00b0 + cos2<\/sup>2019\u00b0
\n= 1 [\u2235 sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1]<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28228.
\nx \u0c12\u0c15 \u0c05\u0c32\u0c4d\u0c2a\u0c15\u0c4b\u0c23\u0c2e\u0c41 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 sin x = cos x \u0c05\u0c2f\u0c3f\u0c24\u0c47 x \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c0e\u0c02\u0c24 ?
\n\u0c1c\u0c35\u0c3e\u0c2c\u0c41 :
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28\u0c26\u0c3f, sin x = cos x
\nsin (90\u00b0- \u03b8) = cos \u03b8 \u0c2e\u0c28\u0c15\u0c41 \u0c24\u0c46\u0c32\u0c41\u0c38\u0c41.
\ncos x = sin(90\u00b0 -x)
\n= sin x = sin(90\u00b0 -x)
\n(\u0c17\u0c2e\u0c28\u0c3f\u0c15 : sin A = sin B, \u0c05\u0c2f\u0c3f\u0c24\u0c47 A = B]
\n\u21d2 x = 90\u00b0 – x
\n\u21d2 2x = 90\u00b0
\n\u2234 x = 45\u00b0<\/p>\n","protected":false},"excerpt":{"rendered":"

Practice the\u00a0AP 10th Class Maths Bits with Answers 11th Lesson \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f on a regular basis so that you can attempt exams with utmost confidence. AP Board 10th Class Maths Bits 11th Lesson \u0c24\u0c4d\u0c30\u0c3f\u0c15\u0c4b\u0c23\u0c2e\u0c3f\u0c24\u0c3f \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c281. \u0c2a\u0c48 \u0c2a\u0c1f\u0c02\u0c32\u0c4b sin C = \u0c05\u0c2f\u0c3f\u0c28 cos C \u0c35\u0c3f\u0c32\u0c41\u0c35. \u0c0e\u0c02\u0c24 ? \u0c1c\u0c35\u0c3e\u0c2c\u0c41 : sin C = BC = 4 . 2. …<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[12],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/29410"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=29410"}],"version-history":[{"count":12,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/29410\/revisions"}],"predecessor-version":[{"id":29498,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/29410\/revisions\/29498"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=29410"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=29410"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=29410"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}