The value of $\tan \left[\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right]$ is
The value of $\tan \left[\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right]$ is
$\frac{17}{6}$
$\frac{16}{7}$
$\frac{6}{17}$
$\frac{7}{16}$
Solution
The value of $\tan \left[\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right]$ is to be found out. Let $\quad \cos ^{-1}\left(\frac{4}{5}\right)=\theta \Rightarrow \cos \theta=\frac{4}{5} \Rightarrow \sin \theta=\frac{3}{5}$
$\therefore \tan \theta=\frac{3}{4} \Rightarrow \theta=\tan ^{-1}\left(\frac{3}{4}\right)$
Hence given expression.
$\begin{array}{l}
=\tan \left[\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right]=\tan \left[\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\left(\frac{3}{4}\right)\left(\frac{2}{3}\right)}\right)\right] \\
=\tan \left[\tan ^{-1}\left(\frac{17}{6}\right)\right]=\frac{17}{6}
\end{array}$