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

Answer: (b)

cosec(75° + θ) - sec(15° - θ)

= cosec(75° + θ) - sec [90° - (75° + θ)]

= cosec (75° + θ) - cosec (75° + θ)

= 0

Answer: (d)

We know:

sin 30° = $1/2$

Value of sin increases 0° to 90°

∴ sin 31° > sin 30° and sin 32° > sin 30°

⇒ sin 31° > $1/2 \text"and sin 32°" > 1/2$

On adding both sides, we get

sin 31° + sin 32°

⇒ $1/2 + 1/2 ⇒ sin 31° = sin 32° > 1$

Answer: (c)

Given, 2 $sec^2$ θ + sec θ - 6 = 0

⇒ 2 $sec^2$ θ + 4 sec θ - 3 sec θ - 6 = 0

⇒ 2 sec θ (sec θ + 2) - 3 (sec θ + 2) = 0

⇒ (2 sec θ - 3) (sec θ + 2) = 0

⇒ sec θ = $3/2$ (∵ sec θ ≠ -2, sec θ > 0)

⇒ cos θ = $2/3$

⇒ sin θ = $√{1 - cos^2 θ} = √{1 - 4/9} = {√5}/3$

∴ cosec θ = $1/{\text"sin θ"}$

= $1/{√5/3} = 3/{√5}$

Answer: (b)

Given that, α + β = 90°.....(i)

According to question,

β = $2/3$ α

∴ β = $2/3 α = 2/3 (90° - β)$ [from equation (i)]

⇒ β = 60° - $2/3$ β ⇒ β = 36°

Answer: (a)

${\text"sin x - cos x + 1"}/{\text"sin x + cos x - 1"}$

= ${(\text"sin x - cos x") + 1}/{(\text"sin x + cos x") - 1} × {(\text"sin x + cos x") + 1}/{(\text"sin x + cos x") + 1}$

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

= ${sin^2 x \text"+ sin x cos x + sin x - cos x sin x -" cos^2 x \text"- cos x + sin x + cos x + 1"}/{sin^2 x + cos^2 x \text"+ 2 sin x cos x - 1"}$

= ${sin^2 x + 2 sin x - cos^2 x + 1}/{\text"1 + 2 sin x cos x - 1"}$

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

= ${sin^2 x \text"+ 2 sin x - 1 +" sin^2 x + 1}/{\text"2 sin x cos x"}$

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

= ${\text"sin x + 1"}/{\text"cos x"}$