The value of $x$, where $x0$ and $ an \left(\sec ^{-1}\left(\frac{1}{x} ight) ight)=\sin \left( an ^{-1} 2 ight)$ is

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

Aicharya · Accepted Answer

Given, $	an \left(\sec ^{-1}\left(\frac{1}{x}ight)ight)=\sin \left(	an ^{-1} 2ight)$ $\begin{array}{r}\Rightarrow 	an \left(	an ^{-1} \frac{\sqrt{1-x^2}}{x}ight)=\sin \left(\sin ^{-1} \frac{2}{\sqrt{1+2^2}}ight) \ \left(\because 	an ^{-1} x=\sin ^{-1} \frac{x}{\sqrt{1+x^2}}ight)\end{array}$ $\begin{aligned} \Rightarrow & & \frac{\sqrt{1-x^2}}{x} & =\frac{2}{\sqrt{5}} \ & \Rightarrow & \sqrt{5} \sqrt{1-x^2} & =2 x \ & \Rightarrow & 4 x^2 & =5\left(1-x^2ight) \ & \Rightarrow & x^2 & =\frac{5}{9} \ & \Rightarrow & x & =\frac{\sqrt{5}}{3}\end{aligned}$

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

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