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

Answer: (a)

We know that,

π radian = 180°

⇒ 1 radian = ${180°}/{π} = {180°}/{22}$ × 7°

= ${630°}/{11} = 57 {3°}/{11} = 57° + {3 × 60}/{11}$ min

= 57° + 16' + $4/{11}$ min

= 57° + 16' + $4/{11}$ × 60 s = 57° + 16' + 21.8''

= 57° + 16' 21.8'' = 57° 16' 22''

Answer: (a)

From question

0 < (θ, φ) < 90°

Graph of cos θ and cos φ:

As cos φ > cos θ

thus θ >φ

Answer: (d)

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

= ${\text"cos θ [1 + sin θ + 1 - sin θ"]}/{(1 - \text"sin θ") (1 + \text"sin θ")}$

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

II. ${\text"cos θ"}/{1 + \text"cosec θ"} + {\text"cos θ"}/{\text"cosec θ" - 1}$

= ${\text"cos θ [cosec θ - 1 + 1 + cosec θ]"}/{\text"cosec"^2 θ - 1}$

= ${\text"2 cos cosec θ"}/{\text"cot"^2 θ} = {\text"2 cot θ"}/{cot^2 θ} = 2/{\text"cot θ"}$

Thus, neither I nor II independent of θ

Answer: (c)

$tan^2 x + 1/{tan^2 x} = 2$

$(tan^2 x + 1/{tan^2 x} - 2)$ = 0

$(\text"tan x - cot x")^2$ = 0

tan x = cot x

when x = 45°

Answer: (d)

We know that in a cyclic quadrilateral sum of opposite angle is 180°.

∴ A + C = 180° .....(i)

and B + D = 180° ......(ii)

∴ cos A + cos B + cos C + cos D

= cos A + cos B + cos (180° - A) + cos (180° - B)

From equations (i) and (ii),

= cos A + cos B - cos A - cos B = 0