sin−1{sin(3π−10)}= 3π−10 [∴3π−10∈[−π2,π2]]

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

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10 - 3 π

3 π - 10

3 π + 10

answer is C.

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Detailed Solution

\sin^{- 1} {\sin (3 π - 10)} =

3 π - 10

$[∴ 3 π - 10 \in [- \frac{π}{2}, \frac{π}{2}]]$

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