Question
Prove that:
$\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}=\tan^{-1}\frac{77}{36}$
$\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}=\tan^{-1}\frac{77}{36}$
✓
Answer
$\text{Let}\sin^{-1}\frac{8}{17}=x.\text{Then}, \sin x=\frac{8}{17}$ $\Rightarrow\cos x=\sqrt{1-\left(\frac{8}{17}\right)^2}=\sqrt{\frac{225}{289}}=\frac{15}{17}.$ $\therefore\tan x=\frac{8}{15}\Rightarrow x=\tan^{-1}\frac{8}{15}$ $\therefore\sin^{-1}\frac{8}{17}=\tan^{-1}\frac{8}{15}\ \dots\dots(1)$ $\text{Now}, \text{let}\sin^{-1}\frac{3}{5}=y.\text{Then},\ \sin y=\frac{3}{5}$ $\Rightarrow\cos y=\sqrt{1-\left(\frac{3}{5}\right)^2}=\sqrt{\frac{16}{25}}=\frac{4}{5}.$ $\therefore\tan y=\frac{3}{4}\Rightarrow y=\tan^{-1}\frac{3}{4}$ $\therefore\sin^{-1}\frac{3}{5}=\tan^{-1}\frac{3}{4} \dots\dots(2)$ Now, we have: $\text{L.H.S.}=\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}$$=\tan^{-1}\frac{8}{15}+\tan^{-1}\frac{3}{4}$ [Using (1) and (2)]
$=\tan^{-1}\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15}\times\frac{3}{4}}$
$=\tan^{-1}\left(\frac{32+45}{60-24}\right)$ $\bigg[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}\bigg]$
$=\tan^{-1}\frac{77}{36}=\text{R.H.S.}$
$=\tan^{-1}\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15}\times\frac{3}{4}}$
$=\tan^{-1}\left(\frac{32+45}{60-24}\right)$ $\bigg[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}\bigg]$
$=\tan^{-1}\frac{77}{36}=\text{R.H.S.}$
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