Question

$tan {}^{ - 1} \frac{1}{4} + 2tan {}^{ - 1} \frac{1}{5} + tan {}^{ - 1} \frac{1}{6} + tan {}^{ - 1} \frac{1}{x} = \frac{\pi}{4}$
${\bf{\green{Solve\:the equation\:for\:x}}}$⬆️⬆️
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Answer 1

AnsWer :

- 461 / 9.

SolutioN :

$\tt \mapsto {tan}^{ - 1} \bigg(\dfrac{1}{4} \bigg) + 2 {tan}^{ - 1}\bigg( \dfrac{1}{5} \bigg)+ {tan}^{ - 1} \bigg(\dfrac{1}{6 }\bigg) + {tan}^{ - 1} \bigg(\dfrac{1}{x}\bigg)=\dfrac{\pi}{4}$

Now, We know

2 tan^-1 x = tan^-1( 2x / 1 - x² )

$\tt \mapsto {tan}^{ - 1} \bigg(\dfrac{1}{4} \bigg) + {tan}^{ - 1}\bigg( \dfrac{ \frac{2}{5} }{1 - \frac{1}{ {5}^{2} } } \bigg)+ {tan}^{ - 1} \bigg(\dfrac{1}{6 }\bigg) + {tan}^{ - 1} \bigg(\dfrac{1}{x}\bigg)=\dfrac{\pi}{4}$

$\tt \mapsto {tan}^{ - 1} \bigg(\dfrac{1}{4} \bigg) + {tan}^{ - 1}\bigg( \dfrac{5}{12} \bigg)+ {tan}^{ - 1} \bigg(\dfrac{1}{6 }\bigg) + {tan}^{ - 1} \bigg(\dfrac{1}{x}\bigg) = {tan}^{ - 1} 1.$

$\tt \mapsto {tan}^{ - 1} \bigg(\dfrac{ \frac{1}{4} + \frac{5}{12} }{1 - \frac{5}{48} } \bigg) + {tan}^{ - 1} \bigg(\dfrac{1}{6 }\bigg) + {tan}^{ - 1} \bigg(\dfrac{1}{x}\bigg) = {tan}^{ - 1} 1.$

$\tt \mapsto {tan}^{ - 1} \bigg( \dfrac{32}{43} \bigg) + {tan}^{ - 1} \bigg(\dfrac{1}{6 }\bigg) + {tan}^{ - 1} \bigg(\dfrac{1}{x}\bigg) = {tan}^{ - 1} 1.$

$\tt \mapsto {tan}^{ - 1} \bigg( \dfrac{192 + 43}{258 - 32} \bigg) + {tan}^{ - 1} \bigg(\dfrac{1}{x}\bigg) = {tan}^{ - 1} 1.$

$\tt \mapsto {tan}^{ - 1} \bigg( \dfrac{235}{226} \bigg) + {tan}^{ - 1} \bigg(\dfrac{1}{x}\bigg) = {tan}^{ - 1} 1.$

$\tt \mapsto {tan}^{ - 1} \bigg(\dfrac{1}{x}\bigg) = {tan}^{ - 1} 1- {tan}^{ - 1} \bigg( \dfrac{235}{226} \bigg)$

$\tt \mapsto {tan}^{ - 1} \bigg(\dfrac{1}{x}\bigg) = {tan}^{ - 1}\bigg(\dfrac{1 - \frac{235}{226} }{1 + \frac{235}{226} }\bigg)$

$\tt \mapsto {tan}^{ - 1} \bigg(\dfrac{1}{x}\bigg) = {tan}^{ - 1}\bigg(\dfrac{ - 9}{461}\bigg)$

$\tt \mapsto \bigg(\dfrac{1}{x}\bigg) = \bigg(\dfrac{ - 9}{461}\bigg)$

$\tt \mapsto x = - \dfrac{461}{9}$

Therefore, the value of x is - 461 / 9.