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

Answer: (b)

Here, $(\text"sin θ + cosec θ")$ = 2.5

⇒ $(sin θ + 1/{\text"sin θ"}) = 5/2$

⇒ $2 sin^2 θ - 5 sin θ + 2 = 0$

⇒ $2 sin^2 θ$ - 4 sin θ - sin θ + 2 = 0

⇒ 2 sin θ (sin θ - 2) - 1 (sin θ - 2) = 0

⇒ (2 sin θ - 1) (sin θ - 2) = 0

⇒ sin θ = $1/2$ (∵ sin θ ≠ 2)

∴ θ = 30°

Answer: (b)

In ΔOAB,

cos 40° = ${AB}/{OB} ⇒ cos 40° = r/R$

⇒ r = R cos 40°

So, the radius of the circle of latitude 40° S is R cos 40°

Answer: (c)

We know that $cos^2$ θ is defined in [0, 1]

so $cos^2 θ = 1 - {p^2 + q^2}/{2pq},$ where p, q are non zero real number, is possible only when p = q.

2. We know that $tan^2$ θ is defined in [0, ∞]. So $tan^2$ θ

= ${4pq}/{(p + q)^2} -1$ where p, q are non zero numbers, is possible only when p = q.

Answer: (a)

∵ cos θ = ${a^2 + b^2 - c^2}/{2ab}$ By cosine rule

= ${6^2 + 2^2 - c^2}/{2 × 6 × 2} = {40 - c^2}/{24}$

For acute angle,

cos θ > 0 ⇒ ${40 - c^2}/{24} > 0 ⇒ c^2 < 40$

⇒ 0 < c < 2 $√{10}$ (since, c cannot be negative) ....(i)

Also, b + c > a

c > 6 - 2 ⇒ c > 4

From equations (i) and (ii),

c ε (4, 2 $√{10}$)

Answer: (a)

Checking statement 1

$(sec^2 θ - 1) (1 - cosec^2 θ) = 1$

L.H.S

$(sec^2 θ - 1) × (- 1) (cosec^2 θ - 1)$

$tan^2 θ × (- 1) × cot^2$ θ = -1

Here L.H.S ≠ R.H.S

Hence, statement 1 is incorrect.

check in statement 2.

sin θ $(1 + cos θ)^{-1} + (1 + cos θ)(sin θ)^{-1}$ = 2 cosec θ

L.H.S

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

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

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

= 2 cosec θ = R.H.S

Hence, only statement 2 is correct.