Q100 / 180 · COMEDK 2019 · Mathematics
$\sin \left(2 \sin ^{-1} \sqrt{\frac{63}{65}}\right)$ is equal to
- Correct answerA$\frac{2 \sqrt{126}}{65}$
- B$\frac{4 \sqrt{65}}{65}$
- C$\frac{8 \sqrt{63}}{65}$
- D$\frac{\sqrt{63}}{65}$
Solution
We have,
$$
\begin{gathered}
\sin \left(2 \sin ^{-1} \sqrt{\frac{63}{65}}\right) \\
=\sin \left\{\sin ^{-1}\left(2 \sqrt{\frac{63}{65}} \sqrt{1-\left(\sqrt{\frac{63}{65}}\right)^{2}}\right)\right\} ; \\
{\left[\because 2 \sin ^{-1} x=\sin ^{-1} 2 x \sqrt{1-x^{2}}\right]}
\end{gathered}
$$
$=\sin \left\{\sin ^{-1}\left(2 \sqrt{\frac{63}{65}} \sqrt{1-\frac{63}{65}}\right)\right\}$
$=\sin \left\{\sin ^{-1}\left(2 \sqrt{\frac{63}{65}} \times \sqrt{\frac{2}{65}}\right)\right\}=\sin \left\{\sin ^{-1} \frac{2 \sqrt{126}}{65}\right\}$
$=\frac{2 \sqrt{126}}{65} \quad\left[\because \sin \left(\sin ^{-1} x\right)=x\right]$
Hence, the correct answer is option: 'A'.