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The value of $x$, where $x>0$ and $\tan \left(\sec ^{-1}\left(\frac{1}{x}\right)\right)=\sin \left(\tan ^{-1} 2\right)$ is
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The correct answer is:
$\frac{\sqrt{5}}{3}$
Given, $\tan \left(\sec ^{-1}\left(\frac{1}{x}\right)\right)=\sin \left(\tan ^{-1} 2\right)$
$\begin{array}{r}\Rightarrow \tan \left(\tan ^{-1} \frac{\sqrt{1-x^2}}{x}\right)=\sin \left(\sin ^{-1} \frac{2}{\sqrt{1+2^2}}\right) \\ \left(\because \tan ^{-1} x=\sin ^{-1} \frac{x}{\sqrt{1+x^2}}\right)\end{array}$
$\begin{aligned} \Rightarrow & & \frac{\sqrt{1-x^2}}{x} & =\frac{2}{\sqrt{5}} \\ & \Rightarrow & \sqrt{5} \sqrt{1-x^2} & =2 x \\ & \Rightarrow & 4 x^2 & =5\left(1-x^2\right) \\ & \Rightarrow & x^2 & =\frac{5}{9} \\ & \Rightarrow & x & =\frac{\sqrt{5}}{3}\end{aligned}$
$\begin{array}{r}\Rightarrow \tan \left(\tan ^{-1} \frac{\sqrt{1-x^2}}{x}\right)=\sin \left(\sin ^{-1} \frac{2}{\sqrt{1+2^2}}\right) \\ \left(\because \tan ^{-1} x=\sin ^{-1} \frac{x}{\sqrt{1+x^2}}\right)\end{array}$
$\begin{aligned} \Rightarrow & & \frac{\sqrt{1-x^2}}{x} & =\frac{2}{\sqrt{5}} \\ & \Rightarrow & \sqrt{5} \sqrt{1-x^2} & =2 x \\ & \Rightarrow & 4 x^2 & =5\left(1-x^2\right) \\ & \Rightarrow & x^2 & =\frac{5}{9} \\ & \Rightarrow & x & =\frac{\sqrt{5}}{3}\end{aligned}$
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