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