Question with Solution

Since ${\cos }^{-1}\left(\frac{4}{5}\right)={\tan }^{-1}\left(\frac{3}{4}\right)$

$$\therefore \tan \left[{\cos }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{2}{3}\right)\right]=\tan \left[{\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{2}{3}\right]=\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\cdot \frac{2}{3}}=\frac{17}{6}$$