sin − 1 ( sin 10 ) = Options: A . &nbsp10 B . &nbsp3π−10 C . &nbspπ+6 D . &nbsp2π−6

Answer: Option B : B 3 π < 10 < 3 π + π 2 ∴ 10 rad ∈ Q 3 ∴ − π 2 < 3 π − 10 < 0 ∴ 3 π − 10 ∈ Q 4 Also sin ( 3 π − 10 ) = sin 10 Hence sin − 1 ( sin 10 ) = sin − 1 sin ( 3 π − 10 ) = 3 π − 10

sin−1(sin10)=

Question

{sin}^{- 1} (sin 10) =

Options:

A . 10

B . 3π−10

C . π+6

D . 2π−6

Answer: Option B
:
B

3 π < 10 < 3 π + \frac{π}{2} ∴ 10 rad \in Q_{3} ∴ - \frac{π}{2} < 3 π - 10 < 0 ∴ 3 π - 10 \in Q_{4} Also sin (3 π - 10) = sin 10 Hence {sin}^{- 1} (sin 10) = {sin}^{- 1} sin (3 π - 10) = 3 π - 10

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