Prove that

    4. \sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5} =\tan^{-1}\frac{77}{36}

Answers (1)

Taking \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 we have-

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

So, we have now,

L.H.S.

\sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5}

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

 =\tan ^{-1} \frac{8}{15} + \tan^{-1} \frac{3}{4}

 =\tan^{-1} \frac{\frac{8}{15}+ \frac{3}{4}}{1-\frac{8}{15}\times \frac{3}{4}}                   [\because \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1 - xy}] 

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

 =\tan^{-1} (\frac{77}{36}) 

=  R.H.S.

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