AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.4

10th Class Maths 11th Lesson Trigonometry Ex 11.4 Textbook Questions and Answers

Question 1.
Evaluate the following:
i) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
Answer:
Given (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Ex 11.4 1

ii) (sin θ + cos θ)2 + (sin θ – cos θ)2
Answer:
Given (sin θ + cos θ)2 + (sin θ – cos θ)2
= (sin2 θ + cos2 θ + 2 sin θ cos θ) + (sin2 θ + cos2 θ – 2 sin θ cos θ) [∵ (a + b)2 = a2 + b2 + 2ab
(a – b)2 = a2 + b2 – 2ab]
= 1 + 2 sin θ cos θ + 1 – 2 sin θ cos θ [∵ sin2 θ + cos2 θ = 1]
= 1 + 1
= 2

iii) (sec2 θ – 1) (cosec2 θ – 1)
Answer:
Given (sec2 θ – 1) (cosec2 θ – 1)
= tan2 θ × cot2 θ [∵ sec2 θ – tan2 θ = 1; cosec2 θ – cot2 θ = 1]
= tan2 θ × \(\frac{1}{\tan ^{2} \theta}\) = 1

AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Ex 11.4

Question 2.
Show that (cosec θ – cot θ)2 = \(\frac{1-\cos \theta}{1+\cos \theta}\)
Answer:
AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Ex 11.4 2

Question 3.
Show that \(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = sec A + tan A
Answer:
Given that L.H.S. = \(\sqrt{\frac{1+\sin A}{1-\sin A}}\)
Rationalise the denominator, rational factor of 1 – sin A is 1 + sin A.
AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Ex 11.4 3
[∵ (a + b)(a + b) = (a + b)2]; (a – b)(a + b) = a2 — b2]
= \(\sqrt{\frac{(1+\sin A)^{2}}{\cos ^{2} A}}\)
= \(\frac{1+\sin A}{\cos A}\)
= \(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\)
= sec A + tan A = R.H.S.

AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Ex 11.4

Question 4.
Show that \(\frac{1-\tan ^{2} A}{\cot ^{2} A-1}\) = tan2 A
Answer:
AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Ex 11.4 4

Question 5.
Show that \(\frac{1}{\cos \theta}\) – cos θ = tan θ – sin θ.
Answer:
AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Ex 11.4 5

AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Ex 11.4

Question 6.
Simplify sec A (1 – sin A) (sec A + tan A)
Answer:
L.H.S. = sec A (1 – sin A) (sec A + tan A)
= (sec A – sec A . sin A) (sec A + tan A)
= (sec A – \(\frac{1}{\cos A}\) . sin A) (sec A + tan A)
= (sec A – tan A) (sec A + tan A)
= sec2 A – tan2 A [∵ sec2 A – tan2 A = 1]
= 1

Question 7.
Prove that (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
Answer:
L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2
= (sin2 A + cosec2 A + 2 sin A . cosec A) + (cos2 A – sec2 A + 2 cos A . sec A) [∵ (a + b)2 = a2 + b2 + 2ab]
= (sin2 A + cos2 A) + cosec2 A + 2 sin A . \(\frac{1}{\sin A}\) + sec2 A + 2 cos A . \(\frac{1}{\cos A}\)
[∵ \(\frac{1}{\sin A}\) = cosec A; \(\frac{1}{\cos A}\) = sec A]
= 1 +(1 + cot2 A) + 2 + (1 + tan2 A) + 2
[∵ sin2 A + cos2 A = 1; cosec2 A = 1 + cot2 A; sec2 A = 1 + tan2 A]
= 7 + tan2 A + cot2 A
= R.H.S.

Question 8.
Simplify (1 – cos θ) (1 + cos θ) (1 + cot2 θ)
Answer:
Given that
(1 – cos θ) (1 + cos θ) (1 + cot2 θ)
= (1 – cos2 θ) (1 + cot2 θ)
[∵ (a – b) (a + b) = a2 – b2]
= sin2 θ. cosec2 θ [∵ 1 – cos2 θ = sin2 θ; 1 + cot2 θ = cosec2 θ]
= sin2 θ . \(\frac{1}{\sin ^{2} \theta}\) [∵ cosec θ = \(\frac{1}{\sin \theta}\)]
= 1

AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Ex 11.4

Question 9.
If sec θ + tan θ = p, then what is the value of sec θ – tan θ?
Answer:
Given that sec θ + tan θ = p ,
We know that sec2 θ – tan2 θ = 1
sec2 θ – tan2 θ = (sec θ + tan θ) (sec θ – tan θ)
= p (sec θ – tan θ)
= 1 (from given)
⇒ sec θ – tan θ = \(\frac{1}{p}\)

Question 10.
If cosec θ + cot θ = k, then prove that cos θ = \(\frac{k^{2}-1}{k^{2}+1}\)
Answer:
Method-I:
Given that cosec θ + cot θ = k
R.H.S. = \(\frac{k^{2}-1}{k^{2}+1}\)
AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Ex 11.4 6

AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Ex 11.4

Method – II:
Given that cosec θ + cot θ = k ……..(1)
We know that cosec2 θ – cot2 θ = 1
⇒ (cosec θ + cot θ) (cosec θ – cot θ) = 1 [∵ a2 – b2 = (a -b)(a + b)]
⇒ k (cosec θ – cot θ) = 1
⇒ (cosec θ – cot θ) = \(\frac{1}{k}\)
By solving (1) and (2)
AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Ex 11.4 7
According to identity cos2 θ + sin2 θ = 1
AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Ex 11.4 8
Hence proved.

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