If sin−1x−cos−1x=π6 , then 'x'=

sin−1x−cos−1x=π6⇒sin−1x−(π2−sin−1x)=π6 2sin−1x=π6+π2⇒2sin−1x=2π3 sin−1x=π3⇒x=sinπ3=32

sin−1x−cos−1x=π6⇒sin−1x−(π2−sin−1x)=π6 2sin−1x=π6+π2⇒2sin−1x=2π3 sin−1x=π3⇒x=sinπ3=32

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If $\sin^{- 1} x - \cos^{- 1} x = \frac{π}{6}$ , then $' x' =$

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\frac{1}{2}

- \frac{1}{2}

\frac{\sqrt{3}}{2}

- \frac{\sqrt{3}}{2}

answer is B.

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\sin^{- 1} x - \cos^{- 1} x = \frac{π}{6} \Rightarrow \sin^{- 1} x - (\frac{π}{2} - \sin^{- 1} x) = \frac{π}{6}

$2 \sin^{- 1} x = \frac{π}{6} + \frac{π}{2} \Rightarrow 2 \sin^{- 1} x = \frac{2 π}{3}$

$\sin^{- 1} x = \frac{π}{3} \Rightarrow x = \sin \frac{π}{3} = \frac{\sqrt{3}}{2}$

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