is equal to:

$2 \pi-\left(\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{16}{65}\right)$

$=2 \pi-\left(\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{16}{63}\right)$ $\left[\because \sin ^{-1} \frac{4}{5}=\tan ^{-1} \frac{4}{3}\right]$

$=2 \pi-\left\{\tan ^{-1}\left(\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3} \cdot \frac{5}{12}}\right)+\tan ^{-1} \frac{16}{63}\right\}$

$=2 \pi-\left(\tan ^{-1} \frac{63}{16}+\tan ^{-1} \frac{16}{63}\right)$

$=2 \pi-\left(\tan ^{-1} \frac{63}{16}+\cot ^{-1} \frac{63}{16}\right)$

$=2 \pi-\frac{\pi}{2}=\frac{3 \pi}{2}$