Question

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

Answer 1

$(ta{n}^{2}1{)}^2+(co{t}^{-1}{)}^2=\frac{5{\pi }^2}{8}$

$(ta{n}^{-1}x{)}^2+(\frac{\pi }{2}-ta{n}^{-1}x{)}^2=\frac{5{\pi }^2}{8}$

$k=ta{n}^{-1}x$

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

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

$(4k-3\pi )(4k+\pi )=0$

$x=tan\frac{3\pi }{4},tan\frac{-\pi }{4}$

$x=-1$