Trigonometric Ratios And Identities Mcq Model Questions & Answers, Practice Test for ibps clerk prelims 2024

Answer: (c)

Required angles of a quadrilateral ABCD are 30°,

60°, 120°, and 150°, respectively.

$sec^2 D - tan^2 D = sec^2 (150°) - tan^2 (150°)$

= $sec^2 (90° + 60°) - tan^2 (90° + 60°)$

= $cosec^2 60° - cot^2$ 60°

= $(2/{√3})^2 - (1/{√3})^2 = 4/3 - 1/3 = 3/3 = 1$

After we know that $sec^2 θ - tan^2 θ = 1$

Similarly, $sec^2 D - tan^2 D$ is always equal to 1.

Answer: (b)

Since, sin x < $1/2,$ 0° < x < 30°

and cos x < $1/2,$ 60° < x < 90°

then, sin x = cos x only for x = 45° in first quadrant.

Hence, option (a) is correct.

Answer: (d)

Complementary angle of 80° = 90° - 80° = 10°

10° can be written as = 10 × ${π}/{180} Rad = {π}/{18}$ rad.

Answer: (d)

(a) sin θ = $√2$ is not possible, since sin θ ≤ 1.

(b) sin θ + cos θ = 2 is not possible, since

- $√2 ≤ sin θ + cos θ ≤ √2.$

(c) sin θ + cos θ = 0

⇒ sin θ = - cos θ ⇒ tan θ = - 1

⇒ tan θ = tan${3 π}/4 ⇒ θ = {3 π}/4$

So, θ does not lie in 0° ≤ θ ≤ 90°.

Thus, option (c) is not correct.

(d) sin θ - cos θ = 1

Squaring both sides,

$sin^2 θ + cos^2 θ$ - 2 sin θ cos θ = 1

∴ 1 - sin 2 θ = 1 ⇒ sin 2 θ = 0 = sin 0°

⇒ θ = ${n π}/2,$ n ε N

θ = 0, ${π}/2$

Thus, option (d) is correct.

Answer: (c)

Statement 1

${\text"cos A"}/{\text"1 - tan A"} + {\text"sin A"}/{\text"1 - cot A"}$

⇒ ${\text"cos A . cos A"}/{\text"cos A - sin A"} + {\text"sin A . sin A"}/{\text"sin A - cos A"}$

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

Statement 2

$(\text"1 - sin A - cos A")^2 = 2 (\text"1 - sin A") (\text"1 + cos A")$

LHS = $(\text"1 - sin A - cos A")^2$

= $1 + sin^2 A + cos^2 A - \text"2 sin A + 2 sin A cos A - 2 cos A"$

= 2 -2 sin A 2 + cos A + 2 sin A cos A

⇒ 2{(1 - sin A) + cos A (1 - sin A)}

= 2 (1 - sin A)(1 + cos A)

So both (1) and (2) are correct.