Prove that: ^ -1 63 16 = ^ -1 5 13 + ^ -1 3 5

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Question

Prove that:
$\tan^{-1}\frac{63}{16}=\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}$

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Answer

$\text{Let}\sin^{-1}\frac{5}{13}=x.\text{Then},\sin x=\frac{5}{13}\Rightarrow\cos x=\frac{12}{13}.$

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

$\therefore\sin^{-1}\frac{5}{13}=\tan^{-1}\frac{5}{12} \dots\dots(1)$

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

$\therefore\tan y=\frac{4}{3}\Rightarrow y=\tan^{-1}\frac{4}{5}.$

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

Using (1) and (2), we have

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

$=\tan^{-1}\frac{5}{12}+\tan^{-1}\frac{4}{3}$

$=\tan^{-1}\Bigg(\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}\times\frac{4}{3}}\Bigg)$ $\left[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}\right]$

$=\tan^{-1}\left(\frac{15+48}{36-20}\right)$

$=\tan^{-1}\frac{63}{16}$

= L.H.S.

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