Prove that

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

Answers (1)

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.

 

 

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