Question

tan invers 1/5+tan invers 1/7+tan invers 1/3+tan invers 1/8 =π\4





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Answer 1

solution:-

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

formula of

$\implies \pink{ \tan {}^{ - 1} x + \tan {}^{ - 1}y = \tan {}^{ - 1} ( \frac{x + y}{1 - xy} ) }$

LHS

$\implies \: \tan {}^{ - 1} ( \frac{ \frac{1}{5} + \frac{1}{7} }{ 1 - \frac{1}{5} .\frac{1}{7} } ) + \tan {}^{ - 1} ( \frac{ \frac{1}{3} + \frac{1}{8} }{1 - \frac{1}{3} . \frac{1}{8} })$

$\implies \: \tan {}^{ - 1} ( \frac{ \frac{ \cancel{12}} { \cancel{35}} }{ \frac{ \cancel{34}}{ \cancel{35}} } ) + \tan {}^{ - 1} ( \frac{ \frac{11}{ \cancel{24}} }{ \frac{23}{ \cancel{24}} } ) \\ \\ \implies \: \tan {}^{ - 1} ( \frac{6}{17} ) + \tan {}^{ - 1} ( \frac{11}{23} ) \\ \\ \implies \: \tan {}^{ - 1}( \frac{ \frac{6}{17} + \frac{11}{23} }{1 - \frac{6}{17} \frac{11}{23} } ) \\ \\ \implies \: \tan {}^{ - 1} \cancel{ ( \frac{ \frac{325}{391} }{ \frac{325}{391} } )}$

$\implies \tan {}^{ - 1} (1) \\ \\ \implies \cancel{\tan {}^{ - 1} ( \tan}( \frac{\pi}{4} ) )$

$\implies \: \frac{\pi}{4}$

RHS =lhs (hence proved)

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