trigonometric ratios and identity Model Questions & Answers, Practice Test for ibps clerk prelims 2023

Answer: (c)

Here, p = a sin x + b cos x and q = a cos x - b sin x

On squaring both sides,

⇒ $p^2 = a^2 sin^2 x + b^2 cos^2 x$ + 2ab sin x cos x ......(i)

q = a cos x - b sin x

On squaring both sides,

and $q^2 = a^2 cos^2 x + b^2 sin^2$ x - 2ab sin x cos x .......(ii)

Now, add equation (i) and equation (ii), we get

∴ $p^2 + q^2 = a^2 (sin^2 x + cos^2 x) + b^2 (cos^2 x + sin^2 x)$

= $a^2 + b^2$

Answer: (c)

x + $1/x$ = 2 cos α

Squaring both sides, then we get

$x^2 + 1/{x^2} + 2 = 4 cos^2 α$

⇒ $x^2 + 1/{x^2} = 2(2 cos^2 α -1)$

=2 $(2 cos^2 α - sin^2 α - cos^2 α)$

= $2 cos^2 α - 2 sin^2 α$

Answer: (c)

We know that, tan θ is increasing in 0° to 90° and tan 45° = 1.

∴ 50° > 1.

So, A is true but R is false.

Answer: (c)

As cot θ = ${63}/{16}$

From Δ ABC

B = 63, P = 16

∴ H = $√{63^2 + 16^2}$ = 65

Now, sin θ + cos θ

= ${P + B}/H = {63 + 16}/{65} = {79}/{65}$

Answer: (a)

Given, α + β = 90°......(i)

∴ $√{\text"cosec α . cosec β"} ({\text"sin α"}/{\text"sin β"} + {\text"cos α"}/{\text"cos β"})^{- 1/2}$

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

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

= $1/{(\text"sin α sin β")^{1/2}} ({\text"sin 90°"}/{\text"cos (90° - α) sin β"})^{- 1/2}$

[from equation (i)]

= $1/{(\text"sin α sin β")^{1/2}} × (\text"sin α sin β")^{1/2}$ = 1