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  • Find the value of 4 tan inverse 1 divided 5 minus tan inverse 1 divided 239

Find the value of  4 \tan^{-1}\frac{1}{5}-tan^{-1}\frac{1}{239}

Answers (1)

We have, 4 tan^{-1}\frac{1}{5}-\tan^{-1}\frac{1}{239}

=2 \times \left ( 2 \tan^{-1}\frac{1}{5} \right )-\tan^{-1}\frac{1}{239}

=2 \left [ \tan^{-1}\frac{\frac{2}{5}}{1-\left ( \frac{1}{5} \right )^{2}} \right ]-\tan^{-1}\frac{1}{239}    \left [ since, 2\tan^{-1}x=\tan^{-1}\frac{2x}{1-\left ( x \right )^{2}} \right ]

 

=2 \left [ \tan^{-1}\frac{\frac{2}{5}}{ \frac{24}{25} } \right ]-\tan^{-1}\frac{1}{239}

=2 \tan^{-1}\frac{5}{12}-\tan^{-1}\frac{1}{239}

=\left [ \tan^{-1}\frac{\frac{5}{6}}{1-\left (\frac{5}{12} \right )^{2}} \right ]-\tan^{-1}\frac{1}{239} \left [ since, 2\tan^{-1}x=\tan^{-1}\frac{2x}{1-\left ( x \right )^{2}} \right ]

=\tan^{-1}\frac{\frac{5}{6}}{1-\frac{25}{144}}-\tan^{-1}\frac{1}{139}

=\tan^{-1}\left ( \frac{144 \times 5}{119 \times 6} \right )-\tan^{-1}\frac{1}{239}

=\tan^{-1}\frac{120}{119}-\tan^{-1}\frac{1}{239}

=\tan^{-1}\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119}.\frac{1}{239}}\left [ since, \tan^{-1} x-\tan^{-1}y=\tan^{-1}\left ( \frac{x-y}{1+xy} \right ) \right ]

=\tan^{-1}\left [\frac{28680-119}{28441+120} \right ]

=\tan^{-1}\frac{28561}{28561}

=\tan^{-1}\left ( 1 \right )

=\tan^{-1}\left ( \tan \frac{\pi}{4} \right )

=\frac{\pi}{4}

Hence, 4 tan^{-1}\frac{1}{5}-\tan^{-1}\frac{1}{239}=\frac{\pi}{4}

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