Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Let ${\tan }^{-1}\left(\frac{4}{3}\right)=\theta \Rightarrow \tan \theta =\frac{4}{3}$

$\therefore \cos \theta =\frac{3}{5} \& \sin \theta =\frac{4}{5}$

So, ${\cos }^{-1}\left\{\frac{3}{10}\cos \left({\tan }^{-1}\frac{4}{3}\right)+\frac{2}{5}\sin \left({\tan }^{-1}\frac{4}{3}\right)\right\}$

$={\cos }^{-1}\left\{\frac{3}{10}\cos \theta +\frac{2}{5}\sin \theta \right\}$

$={\cos }^{-1}\left\{\frac{3}{10}\times \frac{3}{5}+\frac{2}{5}\times \frac{4}{5}\right\}$

$={\cos }^{-1}\left\{\frac{9}{50}+\frac{8}{25}\right\}$

$={\cos }^{-1}\frac{1}{2}=\frac{\pi }{3}$