The value of x where x > 0 and $tan \left(sec^{-1}(\frac{1}{x})\right) = sin (tan^{-1} 2)$, is 

$\frac{\sqrt{5}}{3}$

We have,

$tan \left(sec^{-1}(\frac{1}{x})\right) = sin (tan^{-1} 2)$

$⇒ tan \left(tan^{-1}\frac{\sqrt{1-x^2}}{x}\right) = sin \left( sin^{-1}\frac{2}{\sqrt{5}}\right)$

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

$⇒\frac{1}{x^2}=\frac{9}{5}⇒x =\frac{\sqrt{5}}{3}$                 [∵ x > 0 ]