Inverse Trigonometric Functions

Evaluate $\;$ $\tan \left(\cos^{-1} \dfrac{8}{17}\right)$


Let $\;$ $\cos^{-1} \left(\dfrac{8}{17}\right) = \theta$ $\;\;\; \cdots \; (1)$

Then, $\;$ $\cos \theta = \dfrac{8}{17}$

and $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \dfrac{64}{289}} = \sqrt{\dfrac{225}{289}} = \dfrac{15}{17}$

$\therefore \;$ $\tan \theta = \dfrac{\sin \theta}{\cos \theta} = \dfrac{15}{8}$ $\;\;\; \cdots \; (2)$

Now, $\;$ $\tan \left(\cos^{-1} \dfrac{8}{17}\right) = \tan \theta = \dfrac{15}{8}$ $\;\;\;$ [By equations $(1)$ and $(2)$]