{"id":35015,"date":"2022-08-24T17:41:17","date_gmt":"2022-08-24T12:11:17","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35015"},"modified":"2022-08-24T17:41:17","modified_gmt":"2022-08-24T12:11:17","slug":"inter-1st-year-maths-1a-inverse-trigonometric-functions-solutions-ex-8a","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-1st-year-maths-1a-inverse-trigonometric-functions-solutions-ex-8a\/","title":{"rendered":"Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a)"},"content":{"rendered":"

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions<\/a> Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Exercise 8(a) will help students to clear their doubts quickly.<\/p>\n

Intermediate 1st Year Maths 1A Inverse Trigonometric Functions Solutions Exercise 8(a)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nEvaluate the following.
\n(i) \\(\\sin ^{-1}\\left(\\frac{-\\sqrt{3}}{2}\\right)\\)
\nSolution:
\n\"Inter<\/p>\n

(ii) \\(\\cos ^{-1}\\left(\\frac{1}{\\sqrt{2}}\\right)\\)
\nSolution:
\n\\(\\cos ^{-1}\\left(\\frac{1}{\\sqrt{2}}\\right)=\\cos ^{-1}\\left(\\cos \\frac{\\pi}{4}\\right)=\\frac{\\pi}{4}\\)<\/p>\n

\"Inter<\/p>\n

(iii) sec-1<\/sup>(-\u221a2)
\nSolution:
\n\"Inter<\/p>\n

(iv) cot-1<\/sup>(-\u221a3)
\nSolution:
\n\"Inter<\/p>\n

(v) \\(\\sin \\left(\\frac{\\pi}{3}-\\sin ^{-1}\\left(\\frac{-1}{2}\\right)\\right)\\)
\nSolution:
\n\"Inter<\/p>\n

(vi) \\(\\sin ^{-1}\\left(\\sin \\frac{5 \\pi}{6}\\right)\\)
\nSolution:
\n\"Inter<\/p>\n

(vii) \\(\\cos ^{-1}\\left(\\cos \\frac{5 \\pi}{4}\\right)\\)
\nSolution:
\n\"Inter<\/p>\n

Question 2.
\nFind the values of
\n(i) \\(\\sin \\left(\\cos ^{-1} \\frac{3}{5}\\right)\\)
\nSolution:
\n\\(\\sin \\left(\\cos ^{-1} \\frac{3}{5}\\right)=\\sin \\left(\\sin ^{-1} \\frac{4}{5}\\right)=\\frac{4}{5}\\)<\/p>\n

(ii) \\(\\tan \\left({cosec}^{-1} \\frac{65}{63}\\right)\\)
\nSolution:
\n\\(\\tan \\left({cosec}^{-1} \\frac{65}{63}\\right)=\\tan \\left(\\tan ^{-1} \\frac{63}{16}\\right)\\) = \\(\\frac{63}{16}\\)<\/p>\n

(iii) \\(\\sin \\left(2 \\sin ^{-1} \\frac{4}{5}\\right)\\)
\nSolution:
\n\"Inter<\/p>\n

(iv) \\(\\sin ^{-1}\\left(\\sin \\frac{33 \\pi}{7}\\right)\\)
\nSolution:
\n\"Inter<\/p>\n

(v) \\(\\cos ^{-1}\\left(\\cos \\frac{17 \\pi}{6}\\right)\\)
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 3.
\nSimplify each of the following.
\n(i) \\(\\tan ^{-1}\\left[\\frac{\\sin x}{1+\\cos x}\\right]\\)
\nSolution:
\n\"Inter<\/p>\n

(ii) tan-1<\/sup>(sec x + tan x)
\nSolution:
\n\"Inter<\/p>\n

(iii) \\(\\tan ^{-1} \\sqrt{\\frac{1-\\cos x}{1+\\cos x}}\\)
\nSolution:
\n\"Inter<\/p>\n

(iv) sin-1<\/sup>(2 cos2<\/sup>\u03b8 – 1) + cos-1(1 – 2 sin2<\/sup>\u03b8)
\nSolution:
\nsin-1<\/sup>(cos 2\u03b8) + cos-1<\/sup>(cos 2\u03b8)
\n= sin-1<\/sup>[sin (90\u00b0 – 2\u03b8)] + cos-1<\/sup>(cos 2\u03b8)
\n= 90\u00b0 – 2\u03b8 + 2\u03b8
\n= 90\u00b0<\/p>\n

(v) \\(\\tan ^{-1}\\left(x+\\sqrt{1+x^{2}}\\right)\\); x \u2208 R
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nProve that
\n(i) \\(\\sin ^{-1} \\frac{3}{5}+\\sin ^{-1} \\frac{8}{17}=\\cos ^{-1} \\frac{36}{85}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

(ii) \\(\\sin ^{-1} \\frac{3}{5}+\\cos ^{-1} \\frac{12}{13}=\\cos ^{-1} \\frac{33}{65}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

(iii) \\(\\tan \\left[\\cot ^{-1} 9+{cosec}^{-1} \\frac{\\sqrt{41}}{4}\\right]=1\\)
\nSolution:
\n\"Inter<\/p>\n

(iv) \\(\\cos ^{-1} \\frac{4}{5}+\\sin ^{-1} \\frac{3}{\\sqrt{34}}=\\tan ^{-1} \\frac{27}{11}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 2.
\nFind the values of
\n(i) \\(\\sin \\left(\\cos ^{-1} \\frac{3}{5}+\\cos ^{-1} \\frac{12}{13}\\right)\\)
\nSolution:
\n\"Inter<\/p>\n

(ii) \\(\\tan \\left(\\sin ^{-1} \\frac{3}{5}+\\cos ^{-1} \\frac{5}{\\sqrt{34}}\\right)\\)
\nSolution:
\n\"Inter<\/p>\n

(iii) \\(\\cos \\left(\\sin ^{-1} \\frac{3}{5}+\\sin ^{-1} \\frac{5}{13}\\right)\\)
\nSolution:
\nLet \\(\\sin ^{-1} \\frac{3}{5}\\) = \u03b1 and \\(\\sin ^{-1} \\frac{5}{13}\\) = \u03b2
\n\"Inter<\/p>\n

Question 3.
\nProve that
\n(i) \\(\\cos \\left[2 \\tan ^{-1} \\frac{1}{7}\\right]=\\sin \\left[2 \\tan ^{-1} \\frac{3}{4}\\right]\\)
\nSolution:
\n\"Inter<\/p>\n

(ii) \\(\\tan \\left[2 \\tan ^{-1}\\left(\\frac{\\sqrt{5}-1}{2}\\right)\\right]=2\\)
\nSolution:
\n\"Inter<\/p>\n

(iii) \\(\\cos \\left\\{2\\left[\\tan ^{-1}\\left(\\frac{1}{4}\\right)+\\tan ^{-1}\\left(\\frac{2}{9}\\right)\\right]\\right\\}=\\frac{3}{5}\\)
\nSolution:
\n\"Inter<\/p>\n

Question 4.
\nProve that
\n(i) \\(\\tan ^{-1} \\frac{1}{7}+\\tan ^{-1} \\frac{1}{13}-\\tan ^{-1} \\frac{2}{9}=0\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

(ii) \\(\\tan ^{-1} \\frac{1}{2}+\\tan ^{-1} \\frac{1}{5}+\\tan ^{-1} \\frac{1}{8}=\\frac{\\pi}{4}\\)
\nSolution:
\n\"Inter<\/p>\n

(iii) \\(\\tan ^{-1} \\frac{3}{4}+\\tan ^{-1} \\frac{3}{5}-\\tan ^{-1} \\frac{8}{19}=\\frac{\\pi}{4}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

(iv) \\(\\tan ^{-1} \\frac{1}{7}+\\tan ^{-1} \\frac{1}{8}\\) = \\(\\cot ^{-1} \\frac{201}{43}+\\cot ^{-1} 18\\)
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 5.
\nShow that
\n(i) sec2<\/sup> (tan-1<\/sup> 2) + cosec2<\/sup> (cot-1<\/sup> 2) = 10
\nSolution:
\nLet a = tan-1<\/sup> 2 \u21d2 tan \u03b1 = 2
\nsec2<\/sup> \u03b1 = 1 + tan-1<\/sup> \u03b1 = 1 + 4 = 5
\nLet \u03b2 = cot-1<\/sup> 2 \u21d2 cot \u03b2 = 2
\ncosec2<\/sup> \u03b2 = 1 + cot2<\/sup> \u03b2 = 1 + 4 = 5
\nLHS = sec2<\/sup> (tan-1<\/sup> 2) + cosec2<\/sup> (cot-1<\/sup> 2)
\n= 5 + 5
\n= 10
\n= RHS<\/p>\n

(ii) Find the value of \\(\\left(\\cos ^{-1} \\frac{4}{5}+\\tan ^{-1} \\frac{2}{3}\\right)\\)
\nSolution:
\n\"Inter<\/p>\n

(iii) If sin-1<\/sup> x – cos-1<\/sup> x = \\(\\frac{\\pi}{6}\\) then find x.
\nSolution:
\n\"Inter<\/p>\n

III.<\/span><\/p>\n

Question 1.
\nProve that
\n(i) \\(2 \\sin ^{-1} \\frac{3}{5}-\\cos ^{-1} \\frac{5}{13}=\\cos ^{-1} \\frac{323}{325}\\)
\nSolution:
\n\"Inter<\/p>\n

(ii) \\(\\sin ^{-1} \\frac{4}{5}+2 \\tan ^{-1} \\frac{1}{3}=\\frac{\\pi}{2}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

(iii) \\(4 \\tan ^{-1} \\frac{1}{5}+\\tan ^{-1} \\frac{1}{99}-\\tan ^{-1} \\frac{1}{70}=\\frac{\\pi}{4}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 2.
\n(i) If \u03b1 = \\({tan}^{-1}\\left(\\frac{\\sqrt{1+x^{2}}-\\sqrt{1-x^{2}}}{\\sqrt{1+x^{2}}+\\sqrt{1-x^{2}}}\\right)\\) then prove that x2<\/sup> = sin 2\u03b1.
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

(ii) Prove that tan\\(\\left\\{2-\\tan ^{-1}\\left(\\frac{\\sqrt{1+x^{2}}-1}{x}\\right)\\right\\}=\\mathbf{x}\\)
\nSolution:
\n\"Inter<\/p>\n

(iii) Prove that \\(\\sin \\left[\\cot ^{-1} \\frac{2 x}{1-x^{2}}+\\cos ^{-1}\\left(\\frac{1-x^{2}}{1+x^{2}}\\right)\\right]\\) = 1
\nSolution:
\n\"Inter<\/p>\n

(iv) Prove that \\(\\left\\{\\frac{\\pi}{4}+\\frac{1}{2} \\cos ^{-1}\\left(\\frac{a}{b}\\right)\\right\\}\\)
\nSolution:
\n\"Inter<\/p>\n

Question 3.
\n(i) If cos-1<\/sup> p + cos-1<\/sup> q + cos-1<\/sup> r = \u03c0, then prove that p2<\/sup> + q2<\/sup> + r2<\/sup> + 2pqr = 1
\nSolution:
\nLet cos-1<\/sup> p = A, cos-1<\/sup> q = B and cos-1<\/sup> r = C
\nthen A + B + C = \u03c0 ………(1)
\nand p = cos A, q = cos B and r cos C
\nNow p2<\/sup> + q2<\/sup> + r2<\/sup> = cos2<\/sup> A + cos2<\/sup> B + cos2<\/sup> C
\n= cos2<\/sup> A + (1 – sin2<\/sup> B + cos2<\/sup> C)
\n= 1 + (cos2<\/sup> A – sin2<\/sup> B) + cos2<\/sup> C
\n= 1 + cos (A + B) . cos (A – B) + cos2<\/sup> C
\n= 1 + cos (\u03c0 – C) cos (A – B) + cos2<\/sup> C (By (1))
\n= 1 – cos C cos (A – B) + cos2<\/sup> C
\n= 1 – cos C [cos (A – B) – cos C]
\n= 1 – cos C [cos (A – B) – cos(180\u00b0 – \\(\\overline{A+B}\\)]
\n= 1 – cos C [cos (A – B) + cos (A + B)]
\n= 1 – cos C [2 cos A cos B]
\n= 1 – 2 pqr
\n\u2234 p2<\/sup> + q2<\/sup> + r2<\/sup> + 2pqr = 1<\/p>\n

\"Inter<\/p>\n

(ii) If \\(\\sin ^{-1}\\left(\\frac{2 p}{1+p^{2}}\\right)-\\cos ^{-1}\\left(\\frac{1-q^{2}}{1+q^{2}}\\right)\\) = \\(\\tan ^{-1}\\left(\\frac{2 x}{1-x^{2}}\\right)\\), then prove that x = \\(\\frac{p-q}{1+p q}\\)
\nSolution:
\n\"Inter<\/p>\n

(iii) If a, b, c are distinct non-zero real numbers having the same sign, prove that \\(\\cot ^{-1}\\left(\\frac{a b+1}{a-b}\\right)+\\cot ^{-1}\\left(\\frac{b c+1}{b-c}\\right)\\) + \\(\\cot ^{-1}\\left(\\frac{c a+1}{c-a}\\right)\\) = \u03c0 or 2\u03c0
\nSolution:
\nSince (a – b) + (b – c) + (c – a) = 0.
\n(a – b), (b – c), (c – a) all cannot have the same sign.
\nNow two cases arise, namely, either two of these numbers are positive and one negative (or) two of these numbers are negative and one is positive.
\nCase (i): Without loss of generality, we assume that (a – b), (b – c) are both positive and (c – a) is negative
\n\"Inter
\nCase (ii): Without loss of generality, we assume that (a – b) and (b – c) are both negative and (c – a) is positive.
\n\"Inter<\/p>\n

(iv) If sin-1<\/sup> (x) + sin-1<\/sup> (y) + sin-1<\/sup> (z) = \u03c0, then prove that \\(x \\sqrt{1-x^{2}}+y \\sqrt{1-y^{2}}+z \\sqrt{1-z^{2}}=2 x y z\\)
\nSolution:
\nLet sin-1<\/sup> (x) = A, sin-1<\/sup> (y) = B and sin-1<\/sup> (z) = C
\nThen A + B + C = \u03c0 …………(1)
\nand x = sin A, y = sin B and z = sin C
\nNow LHS = \\(x \\sqrt{1-x^{2}}+y \\sqrt{1-y^{2}}+z \\sqrt{1-z^{2}}\\)
\n= sin A \\(\\sqrt{1-\\sin ^{2} A}\\) + sin B \\(\\sqrt{1-\\sin ^{2} B}\\) + sin C \\(\\sqrt{1-\\sin ^{2} C}\\)
\n= sin A cos A + sin B cos B + sin C cos C
\n= \\(\\frac{1}{2}\\) [sin 2A + sin 2B + sin 2C]
\n= \\(\\frac{1}{2}\\) [2 . sin (A + B) cos (A – B) + sin 2C]
\n= \\(\\frac{1}{2}\\) [2 sin (\u03c0 – c). cos (A – B) + sin 2C]
\n= \\(\\frac{1}{2}\\) [2 sin C cos (A – B) + 2 sin C cos C]
\n= \\(\\frac{1}{2}\\) (2 sin C) [cos (A – B) + cos C]
\n= sin C [cos (A – B) + cos (180\u00b0 – \\(\\overline{A+B}\\)]
\n= sin C [cos (A – B) – cos (A + B)]
\n= sin C [2 sin A sin B]
\n= 2 xyz
\n\u2234 \\(x \\sqrt{1-x^{2}}+y \\sqrt{1-y^{2}}+z \\sqrt{1-z^{2}}=2 x y z\\)<\/p>\n

(v) (a) If tan-1<\/sup> x + tan-1<\/sup> y + tan-1<\/sup> z = \u03c0, then prove that x + y + z = xyz.
\nSolution:
\nLet A = tan-1<\/sup> x, B = tan-1<\/sup> y, C = tan-1<\/sup> z
\ntan A = x, tan B = y, tan C = z
\nGiven A + B + C = \u03c0 ……….(1)
\nA + B = \u03c0 – C
\ntan (A + B) = tan (\u03c0 – C)
\n\\(\\frac{\\tan A+\\tan B}{1-\\tan A \\tan B}\\) = -tan C
\n\\(\\frac{x+y}{1-x y}\\) = -z
\nx + y = -z + xyz
\n\u2234 x + y + z = xyz<\/p>\n

(b) If tan-1<\/sup> x + tan-1<\/sup> y + tan-1<\/sup> z = \\(\\frac{\\pi}{2}\\), then prove that xy + yz + zx = 1.
\nSolution:
\nLet tan-1<\/sup> x = A, tan-1<\/sup> y = B and tan-1<\/sup> z = C
\nThen A + B + C = \\(\\frac{\\pi}{2}\\) …….(1)
\nand x = tan A, y = tan B and z = tan C
\n\u2235 A + B + C = \\(\\frac{\\pi}{2}\\)
\nA + B = \\(\\frac{\\pi}{2}\\) – C
\n\u21d2 tan (A + B) = tan(\\(\\frac{\\pi}{2}\\) – C)
\n\u21d2 \\(\\frac{\\tan A+\\tan B}{1-\\tan A \\cdot \\tan B}\\) = cot C
\n\u21d2 \\(\\frac{x+y}{1-x y}=\\frac{1}{z}\\)
\n\u21d2 zx + yz = 1 – xy (or) xy + yz + zx = 1<\/p>\n

\"Inter<\/p>\n

Question 4.
\nSolve the following equations for x:
\n(i) \\({Tan}^{-1}\\left(\\frac{x-1}{x-2}\\right)+{Tan}^{-1}\\left(\\frac{x+1}{x+2}\\right)=\\frac{\\pi}{4}\\)
\nSolution:
\n\"Inter<\/p>\n

(ii) \\(\\tan ^{-1}\\left(\\frac{1}{2 x+1}\\right)+\\tan ^{-1}\\left(\\frac{1}{4 x+1}\\right)\\) = \\(\\tan ^{-1} \\frac{2}{x^{2}}\\)
\nSolution:
\nGiven that
\n\"Inter
\n\u21d2 x2<\/sup>(3x + 1) = 2x(4x + 3)
\n\u21d2 x [x(3x + 1) – 2(4x + 3)] = 0
\n\u21d2 x = 0 (or) 3x2<\/sup> – 7x – 6 = 0
\n\u21d2 x = 0 (or) 3x2<\/sup> – 9x + 2x – 6 = 0
\n\u21d2 x = 0 (or) 3x(x – 3) + 2(x – 3) = 0
\n\u21d2 x = 0 (or) (3x + 2) (x – 3) = 0
\n\u21d2 x = 0 (or) 3 (or) \\(\\frac{-2}{3}\\)<\/p>\n

(iii) \\(3 \\sin ^{-1}\\left(\\frac{2 x}{1+x^{2}}\\right)-4 \\cos ^{-1}\\left(\\frac{1-x^{2}}{1+x^{2}}\\right)\\) + \\(2 \\tan ^{-1}\\left(\\frac{2 x}{1-x^{2}}\\right)=\\frac{\\pi}{3}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

(iv) sin-1<\/sup>(1 – x) – 2 sin-1<\/sup> x = \\(\\frac{\\pi}{2}\\)
\nSolution:
\nGiven sin-1<\/sup>(1 – x) – 2 sin-1<\/sup>x = \\(\\frac{\\pi}{2}\\)
\nLet sin-1<\/sup>(1 – x) = \u03b1 and sin-1<\/sup>(x) = \u03b2
\nThen sin \u03b1 = 1 – x and sin \u03b2 = x
\ncos \u03b1 = \\(\\sqrt{1-(1-x)^{2}}\\) and cos \u03b2 = \\(\\sqrt{1-x^{2}}\\)
\nNow sin-1<\/sup>(1 – x) – 2 sin-1<\/sup>(x) = \\(\\frac{\\pi}{2}\\)
\n\u21d2 \u03b1 – 2\u03b2 = \\(\\frac{\\pi}{2}\\)
\n\u21d2 \u03b1 = \\(\\frac{\\pi}{2}\\) + 2\u03b2
\n\u21d2 sin \u03b1 = sin (\\(\\frac{\\pi}{2}\\) + 2\u03b2)
\n\u21d2 sin \u03b1 = cos 2\u03b2
\n\u21d2 1 – x = 1 – 2 sin2\u03b2
\n\u21d2 1 – x = 1 – 2x2<\/sup>
\n\u21d2 2x2<\/sup> – x = 0
\n\u21d2 x(2x – 1) = 0
\n\u21d2 x = 0 (or) x = \\(\\frac{1}{2}\\)
\nBut when x = \\(\\frac{1}{2}\\)
\n\"Inter
\nHence x = 0 is the only solution for the given equation.<\/p>\n

\"Inter<\/p>\n

Question 5.
\nSolve the following equations.
\n(i) \\(\\cot ^{-1}\\left(\\frac{1+x}{1-x}\\right)=\\frac{1}{2} \\cot ^{-1}\\left(\\frac{1}{x}\\right)\\), x > 0 and x \u2260 1
\nSolution:
\n\"Inter<\/p>\n

(ii) \\(\\tan \\left[\\cos ^{-1} \\frac{1}{x}\\right]=\\sin \\left[\\cot ^{-1} \\frac{1}{2}\\right]\\); x \u2260 0
\nSolution:
\nLet \\(\\cos ^{-1}\\left(\\frac{1}{x}\\right)=\\alpha, \\cot ^{-1} \\frac{1}{2}=\\beta\\)
\n\"Inter<\/p>\n

(iii) \\(\\cos ^{-1} x+\\sin ^{-1} \\frac{x}{2}=\\frac{\\pi}{6}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

(iv) \\(\\cos ^{-1}(\\sqrt{3} \\cdot x)+\\cos ^{-1} x=\\frac{\\pi}{2}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

(v) \\(\\sin \\left[\\sin ^{-1}\\left(\\frac{1}{5}\\right)+\\cos ^{-1} x\\right]=1\\)
\nSolution:
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Exercise 8(a) will help students to clear their doubts quickly. Intermediate 1st Year Maths 1A Inverse Trigonometric Functions Solutions Exercise 8(a) I. Question 1. Evaluate the following. (i) Solution: (ii) Solution: (iii) sec-1(-\u221a2) Solution: (iv) cot-1(-\u221a3) Solution: …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35015"}],"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\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35015"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35015\/revisions"}],"predecessor-version":[{"id":35016,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35015\/revisions\/35016"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35015"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35015"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35015"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}