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Prove the following:

    3. \tan^{-1}\frac{2}{11} + \tan^{-1}\frac{7}{24} = \tan^{-1}\frac{1}{2}

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Given to prove \tan^{-1}\frac{2}{11} + \tan^{-1}\frac{7}{24} = \tan^{-1}\frac{1}{2}

We have L.H.S

\tan^{-1}\frac{2}{11} + \tan^{-1}\frac{7}{24}

 =\tan^{-1}\frac{\frac{2}{11} + \frac{7}{24} }{1 - \left ( \frac{2}{11}\times\frac{7}{24} \right ) }        \left [ \because \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x +y}{1 - xy} \right ]

=\tan^{-1}\frac{11\times 24 }{\frac{11\times24 -14}{11\times 24} }

=\tan^{-1}\frac{48 + 77}{264 -14}

 =\tan^{-1}\left ( \frac{125}{250}\right ) = \tan^{-1}\left ( \frac{1}{2} \right ) 

=   R.H.S

Hence proved.

Posted by

Divya Prakash Singh

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