Question

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

Answer 1

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

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

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

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

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

$\therefore {\sin }^{-1}\frac{3}{5}={\tan }^{-1}\frac{3}{4}$ ......(2)

Now, we have:
$\text{L.H.S.}={\tan }^{-1}\frac{8}{15}+{\tan }^{-1}\frac{3}{4}$ ...[Using 1 and 2]

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

$={\tan }^{-1}\left(\frac{32+45}{60-24}\right)$

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