Question

Solve ${\left({\tan }^{-1}x\right)}^2+{\left({\tan }^{-1}x\right)}^2=\frac{5{\pi }^2}{8}$

Answer 1

We have,

${\left({\tan }^{-1}x\right)}^2+{\left({\tan }^{-1}x\right)}^2=\frac{5{\pi }^2}{8}$

$\Rightarrow 2{\left({\tan }^{-1}x\right)}^2=\frac{5{\pi }^2}{8}$

$\Rightarrow {\left({\tan }^{-1}x\right)}^2=\frac{5{\pi }^2}{16}$

$\Rightarrow {\tan }^{-1}x=\sqrt{\frac{5{\pi }^2}{16}}$

$\Rightarrow {\tan }^{-1}x=\sqrt{5}\frac{\pi }{4}$

$\Rightarrow x=\tan \frac{\sqrt{5}\pi }{4}$

Hence, this is the answer.