trigonometric ratios and identities mcq Model Questions & Answers, Practice Test for ibps rrb so officer sclae 2 3 single exam 2024

Answer: (a)

cosec (75° + θ) - sec (15° - θ)

- tan (55° + θ) + cot (35° - θ)

⇒ cosec (75° + θ) - cosec [90° - (75° + θ)]

- tan (55° + θ) + tan [90° - (35° - θ)}

⇒ cosec (75° + θ) - cosec (75° + θ)

- tan (55° + θ) + tan (55° + θ) = 0

Answer: (c)

${sin 19°}/{cos 71°} + {cos 73°}/{sin 17°}$

= ${sin 19°}/{cos (90° - 19°)} + {cos 73°}/{sin (90° - 73°)}$

= ${sin 19°}/{sin 19°} + {cos 73°}/{cos 73°}$ = 1 + 1 = 2

Answer: (b)

Given, $tan^2 y cosec^2 x - 1 = tan^2 y$

⇒ $tan^2 y cosec^2 x - tan^2 y$ = 1

⇒ $tan^2 y (cosec^2 x - 1) = 1$

⇒ $tan^2 y . cot^2 x = 1$

⇒ $cot^2 x = cot^2 y$

⇒ x = y

∴ = x - y = 0

Answer: (a)

tan 1° . tan 2° . tan 3° .....tan 87° . tan 88°, tan 89°

tan 1° . tan 2° . tan 3°....tan(90° - 3°). tan (90° - 2°).

tan(90° - 1°)

tan 1° . tan 2° . tan 3°.....cot 3° . cot 1°

tan 1° . tan 2° . tan 3°.....$1/{tan 3°} . 1/{tan 2°} . 1/{tan 1°}$ = 1

Answer: (b)

Both A and R are individually true and R is correct explanation of A.