# Prove

Question:

Prove $\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{77}{36}$

Solution:

Let $\sin ^{-1} \frac{8}{17}=x$. Then, $\sin x=\frac{8}{17} \Rightarrow \cos x=\sqrt{1-\left(\frac{8}{17}\right)^{2}}=\sqrt{\frac{225}{289}}=\frac{15}{17}$.

$\therefore \tan x=\frac{8}{15} \Rightarrow x=\tan ^{-1} \frac{8}{15}$

$\therefore \sin ^{-1} \frac{8}{17}=\tan ^{-1} \frac{8}{15}$ ....(i)

Now, let $\sin ^{-1} \frac{3}{5}=y$. Then, $\sin y=\frac{3}{5} \Rightarrow \cos y=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$.

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

$\therefore \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4}$....(ii)

Now, we have:

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

$=\tan ^{-1} \frac{8}{15}+\tan ^{-1} \frac{3}{4} \quad[U \operatorname{sing}(1)$ and $(2)]$

$=\tan ^{-1} \frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15} \times \frac{3}{4}}$

$=\tan ^{-1}\left(\frac{32+45}{60-24}\right) \quad\left[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}\right]$

$=\tan ^{-1} \frac{77}{36}=$ R.H.S.