Prove that : `sin^(-1)(8/(17))+sin^(-1)(3/5)=sin^(-1)((77)/(85))=tan^(-1)((77)/(36))`

Explanation

Let `\sin ^{-1} (8/17) =x`.


Then `\sin x=\frac{8}{17} \Rightarrow \cos x=\sqrt{1-(\frac{8}{17})^{2}}=\sqrt{\frac{225}{289}}=\frac{15}{17}` `\therefore \tan x=\frac{8}{15} \Rightarrow x=\tan ^{-1} \frac{8}{15}`


`\therefore \sin ^{-1} \frac{8}{17}=\tan ^{-1} \frac{8}{15}`


Now let `\sin ^{-1} \frac{3}{5}=y`. Then `\sin y=\frac{3}{5} \Rightarrow \cos y=\sqrt{1-(\frac{3}{5})^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}` `\therefore \tan y=\frac{3}{4} \Rightarrow \quad y=\tan ^{-1} \frac{3}{4}`


`\therefore \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4} \quad \ldots .2`


Now, we have:


L.H.S `=\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5}`


`=\tan ^{-1} \frac{8}{15}+\tan ^{-1} \frac{3}{4} \quad [`Using 1 and 2 ]


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