If tan(sec^-1x)= sin (cos^-11/(√5)), then x =

We have,

$ tan(sec^{-1}x)= sin (cos^{-1}\frac{1}{\sqrt{5}})$

$⇒ \sqrt{sec^2(sec^{-1}x)-1}= \sqrt{1-cos^2 (cos^{-1}\frac{1}{\sqrt{5}})}$

$⇒ \sqrt{x^2-1} =\sqrt{1-\frac{1}{5}}⇒ x^2 -1 =\frac{4}{5} ⇒ x = ±\frac{3}{\sqrt{5}}$