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

$±\frac{3}{\sqrt{5}}$ $±\frac{\sqrt{5}}{3}$ $±\sqrt{\frac{3}{5}}$ none of these

$±\frac{3}{\sqrt{5}}$

We have, $ tan(sec^{-1}x)= sin (cos^{-1}\frac{1}{\sqrt{5}})$ $⇒ \sqrt{sec^2(sec^{-1}x)-1}= \sqrt{1-cos^2 (cos^{-1}\frac{1}{\sqrt{5}})}$ $⇒ \sqrt{x^2-1} =\sqrt{1-\frac{1}{5}}⇒ x^2 -1 =\frac{4}{5} ⇒ x = ±\frac{3}{\sqrt{5}}$