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

Answer: (a)

Given, p = $tan^2 x + cot^2 x$

= $(\text"tan x + cot x")^2 - 2$

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

= $4/{sin^2 2x}$ - 2

Since, the maximum value of sin 2x is 1.

∴ $P_{min} = 4/1$ - 2 = 2.

∴ p ≥ 2

Hence, p ≥ 2.

Alternate Method

p = $tan^2 x + cot^2 x = tan^2 x + 1/{tan^2 x}$

∴ A.M. ≥ G.M.

∵ $tan^2 x + 1/{tan^2 x} ≥ 2 (tan^2 x. 1/{tan^2 x})^{1/2}$

⇒ $tan^2 x + 1/{tan^2 x}$ ≥ 2 ⇒ P ≥ 2

Answer: (a)

sin x + cos x = c ....(i)

Squaring both sides.

⇒ $sin^2 x + cos^2 \text"x + 2 sinx cos x =" c^2$

⇒ sin x cos = ${c^2 - 1}/2$ ...(ii)

Now, cubing eqn (i) both sides

⇒ $sin^3 x + cos^3 \text"x + 3 sin x cos x (sin x + cos x)" = c^3$

⇒ $sin^3 x + cos^3 x + 3 . {(c^2 - 1)}/2 × c = c^3$

⇒ $sin^3 x + cos^3 x = c^3 - 3/2 (c^2 - 1) c$

⇒ $sin^3 x + cos^3 x = c^3 - {3c^3 + 3c}/2$

$sin^3 x + cos^3 x = {3c - c^3 2}/2$ ...(iii)

On squaring both sides.

⇒ $sin^6 x + cos^6 x + 2 sin^3 x cos^3 x = {(3c - c^3)^2}/4$

⇒ $sin^6 x + cos^6 x + 2 ({(c^2 - 1)}/2)^3 = {9c^2 + c^6 - 6c^4}/4$

⇒ $sin^6 x + cos^6 x$

= ${9c^2 + c^6 - 6c^4 - c^6 + 1 + 3c^2 (c^2 - 1)}/4$

$sin^6 x + cos^6 x = {1 + 6c^2 - 3c^4}/4$

Answer: (d)

Given, $2x^2 cos 60° - 4 cot^2 45° - 2 tan 60°$ = θ

⇒ $2x^2 × 1/2 - 4 (1)^2 - 2 × √3$ = 0

⇒ $x^2 - 4 - 2√3$ = 0

⇒ $x^2 = 4 + 2√3$

⇒ $x^2 = 3 + 1 + 2 √3$

⇒ $x^2 = (√3)^2 + (1)^2 + 2 √3 .1$

⇒ $x^2 = (√3 + 1)^2 ⇒ x = √3 + 1$

Answer: (d)

$sin^2 6° + sin^2 12° + sin^2 18° + ........sin^2 84° + sin^2 90°$

Now, $sin^2 84° = sin^2 (90° - 6°) = cos^2 6°$

∴ $sin^2 6° + sin^2 12° + sin^2 18° + ....cos^2 12° + cos^2 6° + sin^2 90°$

$sin^2 6 + cos^2 6 + ....+ sin^2 42 + cos^2 42 + sin^2 90$

= 7 + 1 = 8

Answer: (d)

sin θ + cos θ = ${1 + √3}/2 = 1/2 + {√3}/2$ i.e., θ = 30°

tan 30 + cot 30 = $1/{√3} + √3 = 4/{√3}$