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

\(\begin{aligned} 2 \pi- & {\left[\left(\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{5}{13}\right)+\sin ^{-1} \frac{16}{65}\right] } \\ & =2 \pi-\left[\left(\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{5}{12}\right)+\sin ^{-1} \frac{16}{65}\right] \\ & =2 \pi-\left[\tan ^{-1}\left(\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3} \times \frac{5}{12}}\right)+\sin ^{-1} \frac{16}{65}\right] \\ & =2 \pi-\left[\tan ^{-1}\left(\frac{63}{16}\right)+\sin ^{-1}\left(\frac{16}{65}\right)\right] \\ & =2 \pi-\left[\cos ^{-1}\left(\frac{16}{65}\right)+\sin ^{-1}\left(\frac{16}{65}\right)\right] \\ & =2 \pi-\frac{\pi}{2}=\frac{3 \pi}{2}\end{aligned}\)