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

Answer: (c)

Since we are given that 0 < x < ${π}/2$, then

sin 0 < x < sin ${π}/2$ ⇒ 0 < sin x < 1 ....(1)

Similarly,

cosec 0 > cosec x > cosec ${π}/2$ ⇒ ∞ > cosec x > 1 ....(2)

Adding (1) and (2), we get

∞ > sin x + cosec x > 2

Thus, we can say that the value of the sum of the two trigonometric functions would be ≥ 2.

Answer: (a)

$√{\text"cos x cosec y - cos x sin y"}$

(∵ x + y = 90°, given)

= $√{cos x . cosec (90° - x) - cos x . sin (90° - x)}$

= $√{\text"cos x . sec x - cos"^2 x}$

= $√{1 - cos^2 x} = √{sin^2 x}$ = sin x

Answer: (d)

sin θ + cosec θ = 2

⇒ sin θ + $1/{\text"sin θ"}$ = 2

⇒ $sin^2$ θ - 2 sin θ + 1 = 0

⇒ $(sin θ - 1)^2$ = 0

⇒ sin θ = 1

⇒ sin θ = sin 90°

⇒ θ = 90°

∴ $sin^4 θ + cos^4 θ = sin^4 90° + cos^4 90°$

= 1 + 0 = 1

Alternate Method

Given that

sin θ + cosec θ = 2

Now, put - θ = 90°

1 + 1 = 2

Similarly put θ = - 90°

$sin^4 θ + cos^4$ θ

= $sin^4 90° + cos^4$ 90°

= 1 + 0 = 1

Answer: (c)

From sin (90 - θ) = cos θ

sin 46° . cos 44° + cos 46° . sin 44°

⇒ sin 46° . sin (90 - 44)° + cos 46° . cos (90 - 44)°

= $sin^2 46° + cos^2 46°$ = 1

Answer: (c)

Given, ${\text"cos x"}/{1 + \text"cosec x"} + {\text"cos x"}/{\text"cosec x" - 1}$ = 2

⇒ ${2 \text"cos x cosec x"}/{cosec^2 x - 1}$ = 2

⇒ ${\text"cos x cosec x"}/{cot^2 x}$ = 1

⇒ tan x = 1

⇒ x = ${π}/4$