class 12 maths inverse trigonometric functions

${\tan ^{ - 1}}\frac{1}{5} + {\tan ^{ - 1}}\frac{1}{7} + {\tan ^{ - 1}}\frac{1}{3} + {\tan ^{ - 1}}\frac{1}{8} = \frac{\pi }{4}$

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📘 Inverse Trigonometric Functions NCERT Misc. , Q.8 , Page 51 SA

${\tan ^{ - 1}}\frac{1}{5} + {\tan ^{ - 1}}\frac{1}{7} + {\tan ^{ - 1}}\frac{1}{3} + {\tan ^{ - 1}}\frac{1}{8} = \frac{\pi }{4}$

Official Solution

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L.H.S.$= \left( {{{\tan }^{ - 1}}\frac{1}{3} + {{\tan }^{ - 1}}\frac{1}{5}} \right) + \left( {{{\tan }^{ - 1}}\frac{1}{7} + {{\tan }^{ - 1}}\frac{1}{8}} \right)$

$= {\tan ^{ - 1}}\left( {\frac{{\frac{1}{3} + \frac{1}{5}}}{{1 - \frac{1}{3} \times \frac{1}{5}}}} \right) + {\tan ^{ - 1}}\left( {\frac{{\frac{1}{7} + \frac{1}{8}}}{{1 - \frac{1}{7} \times \frac{1}{8}}}} \right)$

$= {\tan ^{ - 1}}\left( {\frac{{\frac{8}{{15}}}}{{\frac{{14}}{{15}}}}} \right) + {\tan ^{ - 1}}\left( {\frac{{\frac{{15}}{{56}}}}{{\frac{{55}}{{56}}}}} \right)$

$= {\tan ^{ - 1}}\frac{8}{{14}} + {\tan ^{ - 1}}\frac{{15}}{{55}} = {\tan ^{ - 1}}\frac{3}{{11}}$

$= {\tan ^{ - 1}}\left[ {\frac{{\frac{4}{7} + \frac{3}{{11}}}}{{1 - \frac{4}{7} \times \frac{3}{{11}}}}} \right] = {\tan ^{ - 1}}\left( {\frac{{\frac{{65}}{{77}}}}{{\frac{{65}}{{77}}}}} \right) = {\tan ^{ - 1}}1 = \frac{\pi }{4} = R.H.S.$

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