Question

Prove than tan inverse 1 by 2 + tan inverse 1 by 5 + tan inverse 1 by 8 = pi by 4

Answer 1

$\underline{\bold{To \: prove}}\longrightarrow \\ {tan}^{ - 1} \dfrac{1}{2} + {tan}^{ - 1} \dfrac{1}{5} + {tan}^{ - 1} \dfrac{1}{8} = \dfrac{\pi}{4}$

$\underline{\bold{Concept \: used}}\longrightarrow \\ \boxed{ {tan}^{1} x + {tan}^{ - 1} y = {tan}^{ - 1} \dfrac{x + y}{1 - xy}}$

$\underline{\bold{Solution}}\longrightarrow \\ LHS= {tan}^{ - 1} \dfrac{1}{2} + {tan}^{ - 1} \dfrac{1}{5} + {tan}^{ - 1} \dfrac{1}{8}$

$= ( {tan}^{ - 1} \dfrac{1}{2} + {tan}^{ - 1} \dfrac{1}{5} ) + {tan}^{ - 1} \dfrac{1}{8}$

$= {tan}^{ - 1} \dfrac{ \dfrac{1}{2} + \dfrac{1}{5} }{1 - \dfrac{1}{2} \times \dfrac{1}{5} } \: \: + {tan}^{ - 1} \dfrac{1}{8}$

$= {tan}^{ - 1} \dfrac{ \dfrac{5 + 2}{10} }{ \dfrac{10 - 1}{10} } + {tan}^{ - 1} \dfrac{1}{8}$

$= {tan}^{ - 1} \dfrac{7}{9} + {tan}^{ - 1} \dfrac{1}{8}$

$= {tan}^{ - 1} \dfrac{ \dfrac{7}{9} + \dfrac{1}{8} }{1 - \dfrac{7}{9} \times \dfrac{1}{8} }$

$= {tan}^{ - 1} \dfrac{ \dfrac{56 + 9}{72} }{ \dfrac{72 - 7}{72} }$

$= {tan}^{ - 1} \dfrac{65}{65}$

${tan}^{ - 1} (1)$

$= {tan}^{ - 1} (tan \dfrac{\pi}{4} )$

$= \dfrac{\pi}{4}$=RHS