Answer
$\frac{12}{5}$
Work Step by Step
Given expression is-
$\tan(\sin^{-1}\frac{12}{13})$
Assuming $\sin^{-1} \frac{12}{13}$ = $u$ , we get-
$\sin u$ = $\frac{12}{13}$, [$-\frac{\pi}{2}\leq u\leq \frac{\pi}{2}$]
From First Pythagorean identity-
$\cos u$ = + $\sqrt {1 - \sin^{2} u}$
= $\sqrt {1 - (\frac{12}{13})^{2}} $
= $\sqrt {1 - \frac{144}{169}} $
= $\sqrt { \frac{169 - 144}{169}} $
= $\sqrt { \frac{25}{169}} $ = $\frac{5}{13}$
i.e. $\cos u$ = $\frac{5}{13}$
We know that, $\tan u$ = $\frac{\sin u}{\cos u}$ = $\frac{\frac{12}{13}}{\frac{5}{13}}$ = $\frac{12\times13}{13\times 5}$
i.e. $\tan u$ = $\frac{12}{5}$
i.e. $\tan(\sin^{-1}\frac{12}{13})$ = $\frac{12}{5}$
