trigonometric ratios and identities mcq Model Questions & Answers, Practice Test for ibps clerk prelims 2024

Answer: (b)

(LHS) ${1 + tan^2 θ}/{1 + cot^2 θ} = {sec^2 θ}/{cosec^2 θ}$

= $({\text"sec θ"}/{\text"cosec θ"})^2 = ({\text"sin θ"}/{\text"cos θ"})^2 = tan^2 θ$

R.H.S $({\text"1 - tan θ"}/{\text"1 - cot θ"})^2 = ({\text"1 - tan θ"}/{1 - 1/{\text"tan θ"}})^2$

$[\text"tan θ" ({\text"1 - tan θ"}/{\text"tan θ - 1"})]^2 = (\text"- tan θ")^2$

⇒ $tan^2 θ$

LHS = RHS.

∴ Statement 1 is true when θ ≠ ${π}/9.$

Statement 2

(cot θ) (tan θ) = 1 for all values of θ. Except when θ = 0,90°, 180°.....

∴ Statement 2 is not true.

Option (d) is correct.

Answer: (d)

Given that

sec θ + tan θ = 2

We know that

$sec^2 θ - tan^2 θ = 1$

(sec θ + tan θ) (sec θ - tan θ) = 1

⇒ sec θ - tan θ = $1/2$

Now, adding equations (i) and (ii), we get

2 sec θ = $1/2$ + 2

∴ sec θ = $5/4$

Answer: (b)

From question, we have

3 tan θ = cot θ

3 tan θ = 1/tan θ

$tan^2$ θ = 1/3

tan θ = $1 √3$

θ = π/6

Answer: (c)

sin x - cos x = 0

⇒ sin x = cos x

⇒ x = 45° = ${π}/4$

Now, x = 45° = ${π}/4$

∴ $sin^4 x + cos^4 x = sin^4 {π}/4 + cos^4 {π}/4$

= $(1/{√2})^4 + (1/{√2})^4 = 1/4 + 1/4 = 1/2$

Answer: (a)

We know, cos 90° = 0

∴ cos 1° cos 2° cos 3° ......cos 90° = 0.