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

Answer: (c)

∵ 2 cot θ = 3

⇒ cot θ = $3/2$

∴ ${\text"2 cos θ - sin θ"}/{\text"2 cos θ + sin θ"} = {\text"2 cot θ - 1"}/{\text"2 cot θ + 1"}$

= ${2 × 3/2 - 1}/{2 × 3/2 + 1} = {3 - 1}/{3 + 1} = 2/4 = 1/2$

Answer: (d)

${1 - 2 sin^2 θ cos^2 θ}/{sin^4 θ + cos^4 θ}$

= ${(sin^2 θ + cos^2 θ)^2 - 2 sin^2 θ cos^2 θ}/{sin^4 θ + cos^4 θ}$

= ${sin^4 θ + cos^4 θ}/{sin^4 θ + cos^4 θ}$ = 1

Hence, required sum = 1 + 4 = 5

Answer: (c)

I. ${cot 30° + 1}/{cot 30° - 1}$ = 2 (cos 30° + 1)

⇒ = ${√3 + 1}/{√3 - 1} = 2({√3}/2 + 1)$

⇒ ${√3 + 1}/{√3 - 1} × {√3 + 1}/{√3 + 1} = 2({√3 + 2}/2)$

⇒ ${3 + 1 + 2 √3}/{3 - 1} = √3 + 2$

⇒ ${4 + 2 √3}/2 = √3 + 2$

⇒ ${2 (2 + √3)}/2 = √3 + 2$

⇒ $√3 + 2 = √3 + 2$

Hence, it is true.

II. 2 sin 45° cos 45° - tan 45° cot 45° = 0

⇒ 2 × $(1/{√2} × 1/{√2}) - 1 × 1 = \text"0 or 2" × 1/2 - 1 × 1 = 0$

⇒ 1 - 1 = 0

Hence, both Statements I and II are true.

Answer: (a)

I. Given that, sin x + cos x = 2

⇒ $(sin x + cos x)^2$ = 4

⇒ $(sin^2 x + cos^2 x)$ + 2 sin x cos x = 4

⇒ 1 + sin 2x = 4

⇒ sin 2x = 3

⇒ sin 2x ≠ 3

Hence, there is no value of x in the first quadrant that satisfies sin x + cos x = 2

II. sin x - cos x = 0

⇒ tan x = 1 = tan ${π}/4 ⇒ x = {π}/4$

Also, there is only one value of x in the first quadrant that satisfies sin x - cos x = 0.

Answer: (c)

Given that, $sin^2 x + cos^2 x - 1$ = 0

⇒ $sin^2 x + cos^2 x$ = 1

which is an identify of trigonometric ratio and always true for every real value of x.

Therefore, the equation have an infinite solution.