1. Prove That Sin 12/13 + Cos 4/5+ Tan 63/16= 22/7

if your query is to prove that : ${\sin }^{-1}\frac{12}{13}+{\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{63}{16}=\mathrm{\pi }$
let $A={\sin }^{-1}\frac{12}{13} and B ={\cos }^{-1}\frac{4}{5}$
$$\begin{gathered} \sin A=\frac{12}{13} and \cos B=\frac{4}{5} \\ \Rightarrow \tan A=\frac{12}{5} and \tan B=\frac{3}{4} \\ \Rightarrow A={\tan }^{-1}\frac{12}{5} and B={\tan }^{-1}\frac{3}{4} \\ A+B={\tan }^{-1}\frac{12}{5}+{\tan }^{-1}\frac{3}{4} \\ =\mathrm{\pi }+{\tan }^{-1}\left[\frac{{\displaystyle \frac{12}{5}}+{\displaystyle \frac{3}{4}}}{1-{\displaystyle \frac{12}{5}}*{\displaystyle \frac{3}{4}}}\right] [\sin ce \frac{12}{5}*\frac{3}{4}>1 and both are positive \\ =\mathrm{\pi }+{\tan }^{-1}\left[\frac{48+15}{20-36}\right] \\ =\mathrm{\pi }+{\tan }^{-1}\left[-\frac{63}{16}\right] \\ =\mathrm{\pi }-{\tan }^{-1}\left[\frac{63}{16}\right] \end{gathered}$$

therefore
$$\begin{gathered} A+B+{\tan }^{-1}\frac{63}{16}=\mathrm{\pi } \\ {\sin }^{-1}\frac{12}{13}+{\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{63}{16}=\mathrm{\pi } \end{gathered}$$

which is the required result.

hope this helps you