Prove that: `sin^(-1)(4/5)+sin^(-1)(5/(13))+sin^(-1)((16)/(65))=pi/2`

Explanation

LHS :
$$
\begin{aligned}
&\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{16}{65} \\
&=\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{16}{63} \\
&=\tan ^{-1} \frac{63}{16}+\tan ^{-1} \frac{16}{63} \\
&=\tan ^{-1} \frac{63}{16}+\cot ^{-1} \frac{63}{16} \\
&=\frac{\pi}{2}=\text { RHS }
...