Question

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

Answer 1

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

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

Let ${\tan }^{-1}x=k$

Then $k^2+{\left(\frac{\pi }{2}-k\right)}^2=\frac{5{\pi }^2}{8}$

$k^2+\frac{{\pi }^2}{4}+k^2-\pi k=\frac{5{\pi }^2}{8}$

$=2k^2-\pi k+\frac{{\pi }^2}{4}-\frac{5{\pi }^2}{8}=0$

$=2k^2-\pi k-\frac{3{\pi }^2}{8}=0$

$=16k^2-8\pi k-3{\pi }^2=0$

$=\left(4k-3\pi \right)\left(4k+\pi \right)$

$k=\frac{3\pi }{4},k=\frac{-\pi }{4}$

$x=\tan \left(\frac{3\pi }{4}\right);x=\tan \left(-{\pi }_4\right)$

$x=-1$