Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

We have,

$\tan \left(2{\tan }^{-1}\left(\frac{3}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)\right)$

$=\tan \left({\tan }^{-1}\left(\frac{3}{5}\right)+{\tan }^{-1}\left(\frac{3}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)\right)$

$=\tan \left({\tan }^{-1}\left(\frac{{\displaystyle \frac{3}{5}}+{\displaystyle \frac{3}{5}}}{1-{\displaystyle \frac{9}{25}}}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)\right)$

$=\tan \left({\tan }^{-1}\left(\frac{{\displaystyle 15}}{{\displaystyle 8}}\right)+{\tan }^{-1}\left(\frac{5}{12}\right)\right)$

$={\tan }^{-1}\left[\frac{\frac{15}{8}+\frac{5}{12}}{1-\frac{15}{8}·\frac{5}{12}}\right]={\tan }^{-1}\frac{220}{21}$

$=\tan \left({\tan }^{-1}\frac{220}{21}\right)=\frac{220}{21}$