Prove the following:

    4. 2\tan^{-1} \frac{1}{2} + \tan^{-1}\frac{1}{7} = \tan^{-1}\frac{31}{17}

Answers (1)

Given to prove 2\tan^{-1} \frac{1}{2} + \tan^{-1}\frac{1}{7} = \tan^{-1}\frac{31}{17}

Then taking L.H.S.

We have 2\tan^{-1} \frac{1}{2} + \tan^{-1}\frac{1}{7}

=\tan^{-1} \frac{2.\frac{1}{2}}{1 - \left ( \frac{1}{2} \right )^2} + \tan^{-1} \frac{1}{7}                    \because 2\tan^{-1} x = \tan^{-1} \frac{2x}{1- x^2}

=\tan^{-1} \frac{1}{(\frac{3}{4})} + \tan^{-1} \frac{1}{7}

=\tan^{-1} \frac{4}{3} + \tan^{-1} \frac{1}{7}

=\tan^{-1} \frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3}.\frac{1}{7}}                                  \left [ \because \tan^{-1}x + \tan^{-1} y = \tan^{-1} \frac{x +y}{1- xy}\right ]

 =\tan^{-1} \left ( \frac{\frac{28+3}{21}}{\frac{21-4}{21}} \right )

=\tan^{-1} \frac{31}{17} 

=  R.H.S.

Hence proved.

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