Question

If $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} x$, then $x$ is equal to

• A. $\frac{24}{5}$
• B. $\frac{24}{7}$
• C. $\frac{3}{4}$
• D. $\frac{24}{25}$

Answer 1

Answer: (B)

Let, $ 2 \sin ^{-1} \frac{3}{5}=\sin ^{-1}\left[2 \times \frac{3}{5} \sqrt{1-\left(\frac{3}{5}\right)^2}\right] $
$ {\left[\because 2 \sin ^{-1} y=\sin ^{-1}\left(2 y \sqrt{1-y^2}\right)\right] } $
$= \sin ^{-1}\left[2 \times \frac{3}{5} \times \frac{4}{5}\right]=\sin ^{-1}\left(\frac{24}{25}\right)$
$=\tan ^{-1}\left[\frac{\frac{24}{25}}{\sqrt{1-\left(\frac{24}{25}\right)^2}}\right] \left(\because \sin ^{-1} y=\tan ^{-1} \frac{y}{\sqrt{1-y^2}}\right)$
$=\tan ^{-1}\left[\frac{\frac{24}{25}}{\sqrt{1-\frac{576}{625}}}\right]=\tan ^{-1}\left[\frac{24}{25} \times \frac{25}{7}\right]=\tan ^{-1}\left[\frac{24}{7}\right]$
Hence, $x=\frac{24}{7}$