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

Answer: (d)

As sin θ = $3/5 ⇒ cot θ = 4/3$

$1 + 3x + 9x^2 + 27x^3 + 81x^4 + 243x^5$

= 1 + 3 × $4/3 + 9 × {16}/9 + 27 × {64}/{27} + 81 × {256}/{81} + 243 × {1024}/{243}$

= 1 + 4 + 16 + 64 + 256 + 1024

= 5 + 30 + 1280 = 1365

Answer: (c)

I. R.H.S = $cos^2 θ (1 + tan θ) (1 - tan θ)$

= $cos^2 θ (1 - tan^2 θ)$

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

= ${cos^2 θ - sin^2 θ}/{cos^2 θ + sin^2 θ}$ = L.H.S.

II. L.H.S = ${1 + sin θ}/{1 - sin θ} = {(1 + sin θ)^2}/{1 - sin^2 θ}$

= $({1 + sin θ}/{cos θ})^2 = (sec θ + tan θ)^2$

Hence, both statements are correct.

Answer: (b)

We know that, $sin^2 θ + cos^2 θ$ = 1

I. $sin^2 1° + cos^2$ 1° = 1. It is true.

II. $sec^2 33° - cot^2 57° = cosec^2 37° - tan^2 53°$

Now, $sec^2 (90° - 57°) = cosec^2$ 57°

and $cot^2 57° = cot^2 (90° - 33°) = tan^2 33°$

∴ $sec^2 33° - cot^2 57° = cosec^2 57° - tan^2 33°$

Thus, Statement II is incorrect.

Answer: (d)

Since, tan θ = $3/4 = P/B$

∴ H = $√{P^2 + B^2} = √{9 + 16} = √{25}$ = 5

Let the length of hypotenuse = x cm

∴ sin θ = $2/x = 3/5 ⇒ x = {2 × 5}/2 = {10}/3$ cm

Answer: (c)

1. ${\text"cos θ"}/{\text"1 - sin θ"} + {\text"cos θ"}/{\text"1 + sin θ"}$ = 4

⇒ ${\text"cos θ" (\text"1 + sin θ") \text"+ cos θ" (\text"1 - sin θ")}/{1 - sin^2 θ}$ = 4

⇒ ${\text"cos θ + sin θ cos θ + cos θ - sin θ cos θ"}/{cos^2 θ}$ = 4

⇒ $2/{\text"cos θ"}$ = 4

⇒ sec θ = 2

⇒ sec θ = sec 60°

⇒ θ = 60°

2. 3 tan θ + cot θ = 5 cosec θ

$3/{\text"cot θ"}$ + cot θ = 5 cosec θ

3 + $cot^2$ θ = 5 cosec θ cot θ

2 + $cosec^2$ θ = 5 cosec θ cot θ

$2sin^2$ θ + 1 = 5 cos θ

2 - 2 $cos^2$ θ +1 = 5 cos θ.

2 $cos^2$ θ + 5 cos θ - 3 = 0.

Solving quadratic equation for cos θ, we get

cos θ = $1/2,$ cos θ = -3 (neglecting)

⇒ cos θ = $1/2$ = cos 60°

⇒ θ = 60°