Question
If $\sin ^{-1} x=\frac{\pi}{3}$ then write the value of $\cos ^{-1} x$
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Answer
$\sin ^{-1}\left\{2 x \sqrt{1-x^2}\right\}=\sin ^{-1}\left\{2 \sin \left(\frac{1}{2}\right) \sqrt{1-\sin ^2\left(\frac{1}{2}\right)}\right\}$$
\begin{array}{l}
{\left[\because \text { given } \sin ^{-1}(x)=\frac{1}{2}\right]} \\
=\sin ^{-1}\left\{2 \sin \left(\frac{1}{2}\right) \sqrt{\cos ^2\left(\frac{1}{2}\right)}\right\} \\
=\sin ^{-1}\left\{2 \sin \left(\frac{1}{2}\right) \cos \left(\frac{1}{2}\right)\right\} \\
=\sin ^{-1}\left\{\sin \left(2 \times \frac{1}{2}\right)\right\} \\
=\sin ^{-1}(\sin (1)) \\
=1 \text {}
\end{array}
$
\begin{array}{l}
{\left[\because \text { given } \sin ^{-1}(x)=\frac{1}{2}\right]} \\
=\sin ^{-1}\left\{2 \sin \left(\frac{1}{2}\right) \sqrt{\cos ^2\left(\frac{1}{2}\right)}\right\} \\
=\sin ^{-1}\left\{2 \sin \left(\frac{1}{2}\right) \cos \left(\frac{1}{2}\right)\right\} \\
=\sin ^{-1}\left\{\sin \left(2 \times \frac{1}{2}\right)\right\} \\
=\sin ^{-1}(\sin (1)) \\
=1 \text {}
\end{array}
$
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