Prove that : `tan^(-1)((63)/(16))=sin^(-1)(5/(13))+cos^(-1)(3/5)`

Answer

Proved

Explanation

Let `\sin^{-1}( 5/13)=x`. Then, `\sin x=\frac{5}{13} \Rightarrow \cos x=\frac{12}{13}`


`\therefore \tan x=\frac{5}{12} \Rightarrow x=\tan ^{-1} \frac{5}{12}`


`\therefore \sin ^{-1} \frac{5}{13}=\tan ^{-1} \frac{5}{12} \text {..(1) }`


Let `\frac{\cos ^{-1} 3}{5}=y`. Then `\cos y=\frac{3}{5} \Rightarrow \sin y=\frac{4}{5}`


`\therefore \tan y=\frac{4}{3} \Rightarrow y=\tan ^{-1} \frac{4}{3}`


`\therefore \frac{\cos ^{-1} 3}{5}=\tan ^{-1} \frac{4}{3} \ldots(2)`


Using (1) and (2), we have

R.H.S `=\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}`


...