The value of $x$ where $d>0$ and $\tan[\sec^{-1}\big(\large\frac{1}{x}\big)]=$$\sin(\tan^{-1}2) is (a)\;\sqrt 5\qquad(b)\;\large\frac{\sqrt 5}{3}$$\qquad(c)\;1\qquad(d)\;\large\frac{2}{3}$

$\tan(\sec^{-1}\big(\large\frac{1}{x}\big))$$=\sin(\tan^{-1}2) \Rightarrow \tan\big[\tan^{-1}\large\frac{\sqrt{1-x^2}}{x}\big]=$$\sin\big[\sin^{-1}\large\frac{2}{\sqrt{1+2^2}}\big]$
$\tan^{-1}x=\sin^{-1}\large\frac{x}{\sqrt{1+x^2}}$
$\large\frac{\sqrt{1-x^2}}{x}=\frac{2}{\sqrt 5}$
$\sqrt 5\sqrt{1-x^2}=2x$
$4x^2=5(1-x^2)$
$x^2=\large\frac{ 5}{3}$
$x=\large\frac{\sqrt 5}{3}$
Hence (b) is the correct answer.
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