trigonometric ratios and identity Model Questions & Answers, Practice Test for ssc mts paper 1 2023

Answer: (c)

Given, sin (x - y) = $1/2$ and cos (x + y) = $1/2$

⇒ sin (x - y) = sin 30°

and cos (x + y) = cos 60°

⇒ x - y = 30° and x + y = 60°

∴ x = 45° and y = 15°

Answer: (d)

Given, sin (x + 54°) = cos x

⇒ sin (x + 54°) = sin (90° - x) (∵ 0° < x < 90°)

⇒ x + 54° = 90° - x

⇒ 2x = 36° ⇒ x = 18°

Answer: (c)

A. It is true.

R. We know that, cos θ > sin θ, 0° < θ < 45° and sin θ > cos θ

45° < θ < 90°.

Therefore, A is true but R is false.

Answer: (c)

tan (A + B) = $√3$

tan (A + B) = tan 60°

A + B = 60° ....(i)

Now, tan A = 1

tan A = tan 45°

A = 45°

Now putting the value of A in eqn (i)

B = 60° - 45° = 15°

tan (A - B) = tan (45° - 15°) = tan 30° = $1/{√3}$

Answer: (d)

(cosec x - sin x) (sec x - cos x) (tan x + cot x)

$(1/{\text"sin x"} \text"- sin x") (1/{\text"cos x"} \text"- cos x") ({\text"sin x"}/{\text"cos x"} + {\text"cos x"}/{\text"sin x"})$

= ${(1 - sin^2 x) (1 - cos^2 x) (sin^2 x + cos^2 x)}/{\text"sin x cos x . sin x cos x"}$

= ${cos^2 x sin^2 x × 1}/{sin^2 x cos^2 x}$ = 1