Prove that

    3. 2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}

Answers (1)

To prove: 2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7};

L.H.S=2\sin^{-1}\frac{3}{5}

Assume that  \sin^{-1}\frac{3}{5} = x 

then we have \sin x = \frac{3}{5}.

or \cos x = \sqrt{1-\left (\frac{3}{5} \right )^2} = \frac{4}{5}

Therefore we have

 \tan x = \frac{3}{4}\:\:or\:\:x = \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{3}{4}

Now,

We can write L.H.S as

 2\sin^{-1}\frac{3}{5} = 2\tan^{-1}\frac{3}{4}

 =\tan^{-1} \left [\frac{2\times\frac{3}{4}}{1- \left ( \frac{3}{4} \right )^2} \right ]    as we know    \left [2\tan^{-1} x = \tan^{-1} \frac{2x}{1-x^2} \right ]

 =\tan^{-1} \left [\frac{\frac{3}{2}}{\left ( \frac{16-9}{16} \right )} \right ] = \tan^{-1}\left ( \frac{3}{2}\times \frac{16}{7} \right )

 =\tan^{-1} \frac{24}{7}=R.H.S  

L.H.S = R.H.S

 

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