trigonometric ratios and identities Model Questions & Answers, Practice Test for ssc cgl tier 1 2024

Answer: (b)

When $θ_1 = θ_2 = θ_3 = 0$ then $cos θ_1 + cos θ_2 + cos θ_3$ = 3 then $sin θ_1 + sin θ_2 + sin θ_3$ = 0

Answer: (c)

Given ${AB}/{BC} = 3/4$

sin A + sin B = sin C

= $4/5 + 1 + 3/5 = {4 + 3 +5}/5 = {12}/5$

Answer: (d)

For 0 ≤ x ≤ ${π}/2, cos^2 \text"x and" sin^2 x$ lies between 0 to 1.

Hence, atleast one of $sin^2 x, cos^2 x$ is less than 1.

Answer: (d)

3 - $tan^2 θ = α (1 - 3 tan^2 θ)$

⇒ α = ${3 - tan^2 θ}/{1 - 3 tan^2 θ}$

⇒ α = ${4 - 4 tan^2 θ}/{2 - 2 tan^2 θ}$ [By componendo and dividendo]

= $2({1 - tan^2 θ}/{1 + tan^2 θ})$

⇒ α = 2 cos 2 θ

Now - 1 ≤ cos 2 θ ≤ 1 ⇒ - 2 ≤ 2 cos 2 θ ≤ 2

Answer: (d)

(3 sin θ + 5 cos θ) = 4

Squaring both sides,

$9 sin^2 θ + 25 cos^2 \text"θ + 30 cos θ sin θ"$ = 16

$9 sin^2 θ + 9 cos^2 \text"θ + 16" cos^2 \text"θ + 30 cos θ"$ sin θ = 16.

9 + $16cos^2 θ$ + 30 cos θ sin θ = 16.

30 cos θ sin θ = 16 - 9 - 16 $cos^2$ θ ...(1)

Now

$(\text"3 cos θ - 5 sin θ")^2$

= $9 cos^2 θ + 25 sin^2 \text"θ - 30 sin θ cos θ"$

= $9 cos^2 θ + 9 sin^2 \text"θ + 16" sin^2 θ$ - 30 sin θ cos θ

= 9 + 16 $sin^2 \text"θ - 16 + 9 + 16" cos^2 θ$ [from (1)]

= 18 + 16 - 16 = 18