Gujarat Board GSEB Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4

Question 1.
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. (CBSE 2012)
Solution:

tan A = (frac {1}{cot A})

Question 2.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution:
sin A = (sqrt{sin ^{2} A}) = (sqrt{1-cos ^{2} A})

cos A = (frac {1}{sec A})
tan A = (sqrt{tan ^{2} A}) = (sqrt{sec ^{2} A-1}) ……..(2)

Question 3.
(i) (frac{sin ^{2} 63^{circ}+sin ^{2} 27^{circ}}{cos ^{2} 17^{circ}+cos ^{2} 73^{circ}})
(ii) sin 25° cos 65° + cos 25° sin 65°
Solution:
(i) (frac{sin ^{2} 63^{circ}+sin ^{2} 27^{circ}}{cos ^{2} 17^{circ}+cos ^{2} cdot 73^{circ}})
= (frac{sin ^{2}left(90^{circ}-27^{circ}right)+sin ^{2} 27^{circ}}{cos ^{2} 17+cos ^{2}left(90^{circ}-17^{circ}right)})
= (frac{cos ^{2} 27^{circ}+sin ^{2} 27^{circ}}{cos ^{2} 17^{circ}+sin ^{2} 17^{circ}}=frac{1}{1})
= 1

(ii) sin 25° cog 65° + cos 25° sin 65°
= sin 25° cos (90° – 25°) + cos 25° sin (90° – 25°)
= sin 25° sin 25° + cos 25° cos 25°
= sin² 25° + cos² 25
= 1

Question 4.
Choose the correct option. Justify your choice.
(i) 9 sec² θ – 9 tan² θ =
(a) 1
(b) 9
(c) 8
(d) 0

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)=
(a) –
(b) 1
(c) 2
(d) -1

(iii) (sec A + tan A) (1 – sin A) =
(a) sec A
(b) sin A
(c) cosec A
(d) cos A

(iv) (frac{1+tan ^{2} A}{1+cot ^{2} A}=1) = 1
(a) sec 2A
(b) -1
(c) cot 2A
(d) tan 2A
Solution:
(i) 9 sec 2A – 9 tan 2A
= 9 (sec 2A – tan 2A) = 9 x 1 = 9

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)

Hence, correct answer is (c) 2.

(iii) (sec A + tan A) (1 – sin A)
= (left[frac{1}{cos A}+frac{sin A}{cos A}right]) [ 1 – sin A]
= (left[frac{1+sin mathrm{A}}{cos mathrm{A}}right]) [ 1 – sin A]
= (frac{1-sin ^{2} A}{cos A})
= (frac{cos ^{2} A}{cos A}) = cos A
Hence, correct answer is (d) cos A.

(iv) (frac{1+tan ^{2} A}{1+cot ^{2} A}) = 1

= tan² A
Hence, the correct answer is (d) tan² A.

Question 5.
Prove that the following identities, where the angles involved are acute angles for which the expression are defined.
(i) (cosec θ – cot θ)² = (frac{1-cos theta}{1+cot theta})
(ii) (frac{cos A}{1+sin A}) + (frac{1+sin A}{cos A}) = 2 sec A
(iii) (frac{cos A}{1+sin A}) + (frac{cos A}{1+sin A}) = 1 + sec θ cosec θ
(Hint: write the expression in terms of sin θ and cos θ)
(iv) (frac{cos A}{1+sin A}) + (frac{cos A}{1+sin A})
(Hint: simplify LHS and RHS separately)
(v) (frac{cos A-sin A+1}{cos A+sin A-1}) = cosec A + cot A
using identitify cosec² = 1 + cot² A
(vi) (sqrt{frac{1+sin A}{1-sin A}}) = sec A + tan A
(vii) (frac{sin theta-2 sin ^{3} theta}{2 cos ^{3} theta-cos theta}) = tan θ
(viii) (sin A + cosec A)² + (cot A + sec A)²
= 7 + tan² A + cot² A (CBSE 2012)
(ix) (cosec A – sin A)(sec A – cos A)
= (frac{1}{tan A+cot A})
(Hint: simplify LHS and RHS separately)
(x) (left[frac{1+tan ^{2} A}{1+cot ^{2} A}right]) = (left[frac{1-tan A}{1-cot A}right]^{2})
Solution:
(i) We have
(cosec θ – cot θ)² = (frac{1-cos theta}{1+cot theta})
LHS = (cosec θ – cot θ)²

(ii) We have

(viii) (sin A + cosec A)² + (cot A + see A)²
= 7 + tan² A + cot² A
LHS = (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
= sin²A + cos²A + 1 + cot² A + 2 sin A x
(frac{1}{sin A})+ 1+ tan² A + 2cos A x (frac{1}{cos A})
= 1 + 1 + cot²A + 2 + 1 + tan²A + 2
= 7 + tan²A + cot²A = RHS

(ix) (cosee A – sin A) (sec A – cos A)
= (frac{1}{tan A+cot A})
LHS = (cosee A – sin A) (sec A – cos A)
= (left(frac{1}{sin A}-sin Aright)) (left(frac{1}{cos A}-cos Aright))
= (left(frac{1-sin ^{2} A}{sin A}right)) (left(frac{1-cos ^{2} A}{cos A}right))