Evaluate

We know that

$\tan ^{-1} x=\sin ^{-1} \frac{x}{\sqrt{1+x^{2}}}$

$\therefore \sin \left(\tan ^{-1} \frac{3}{4}\right)=\sin \left\{\sin ^{-1}\left(\frac{\frac{3}{4}}{\sqrt{1+\frac{9}{16}}}\right)\right\}$

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

$=\sin \left(\sin ^{-1} \frac{3}{5}\right)$

$=\frac{3}{5} \quad\left[\because \sin \left(\sin ^{-1} x\right)=x\right]$

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