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Prove that

7. $\tan^{-1}\frac{63}{16} = \sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5}$

Answers (1)

best_answer

Taking R.H.S;

We have $\sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5}$

Converting sin and cos terms in tan forms:

Let $\sin^{-1}\frac{5}{13} = x$ and $\cos^{-1}\frac{3}{5} = y$

now, we have $\sin^{-1}\frac{5}{13} = x$ or $\sin x = \frac{5}{13}$

$\sin x = \frac{5}{13} \:or\: \cos x =\frac{12}{13}\:or\:\tan x = \frac{5}{12}$

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

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

Now, $\cos^{-1}\frac{3}{5} = y\Rightarrow \cos y = \frac{3}{5}$

$\cos y = \frac{3}{5} \:or\: \sin y = \frac{4}{5}\:or\:\tan y = \frac{4}{3}$

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

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

Now, using (1) and (2) we get,

R.H.S.

$\sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5} = \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{4}{3}$

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

so,

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

equal to L.H.S

Hence proved.

Posted by

Divya Prakash Singh

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