Trigonometric Ratios & Identity Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (d)

Cos A = tan B

Squaring on both sides

$cos^2 A = tan^2 B$

⇒ $tan^2 B = {sin^2 B}/{cos^2 B} = {1 - cos^2 B}/{cos^2 B}$

∴ $cos^2 A = {1 - cos^2 B}/{cos^2 B}$

$cos^2 A = {1 - tan^2 C}/{tan^2 C}$

(∵ cos B = tan C)

⇒ $cos^2 A tan^2 C = 1 - tan^2 C = 1 - {sin^2 C}/{cos^2 C} $

⇒ $cos^2 A {sin^2 C}/{cos^2 C} ={cos^2 C - sin^2 C}/{cos^2 C}$

⇒ $cos^2 A (1 - cos^2 C) = 2 cos^2 C - 1$

⇒ $cos^2 A (1 - tan^2 A) = 2 tan^2 A - 1$

$cos^2 A - sin^2 A = {2 sin^2 A}/{cos^2 A} - 1$

⇒ 1 - $2sin^2 A = {2 sin^2 A - cos^2 A}/{cos^2 A}$

⇒ $cos^2 A (1 - sin^2 A) = 2 sin^2 A - cos^2 A$

⇒ $cos^2 A (1 - 2 sin^2 A) = 2 sin^2 A - 1 + sin^2 A$

⇒ $(1 - sin^2 A) (1 - 2 sin^2 A) = 3 sin^2 A - 1$

⇒ $1 - 2 sin^2 A - sin^2 A + 2 sin^4A = 3 sin^2 A - 1$

⇒ $1 - 3 sin^2 A + 2 sin^4 A = 3 sin^2 A - 1$

⇒ $2 sin^4 A - 6 sin^2 A + 2 = 0$

⇒ $sin^4 A - 3 sin^2$ A + 1 = 0

This is quadratic equation in $sin^2$ A

$(sin^2 A)^2 - 3 (sin^2 A) + 1 = 0$

$sin^2 A = {3 ± √{(- 3)^2 - 4 (1) (1)}}/2$

= ${3 ± √5}/2$

$sin^2 A = {3 + √5}/2$ not possible because $sin^2 A$ ≯ 1

So $sin^2 A = {3 - √5}/2$

So none of the options are correct.

Answer: (a)

Given, $\text"sin θ + cos θ"$ = 1

Squaring both sides,

$(sin^2 θ + cos^2 θ) + 2 \text"sin θ cos θ" = 1$

⇒ 1 + 2 sin θ cos θ = 1 ⇒ sin θ cos θ = 0.

Answer: (b)

Given that, cos θ + $√3$ sin θ = 2

⇒ $1/2 cos θ + {√3}/2$ sin θ = 1

⇒ sin 30° cos θ + cos 30° sin θ = 1

⇒ sin(30° + θ) = sin 90°

30° + θ = 90°

∴ θ = 60°