The value of x,where x0 and tan left(sec ^{- | vedclass

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(B) Given equation: $	an \left(\sec ^{-1}\left(\frac{1}{x}ight)ight)=\sin \left(	an ^{-1} 2ight)$.First,simplify the left side: Let $	heta =

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

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(B) Given equation: $	an \left(\sec ^{-1}\left(\frac{1}{x}ight)ight)=\sin \left(	an ^{-1} 2ight)$.First,simplify the left side: Let $	heta = \sec^{-1}(\frac{1}{x})$,then $\sec 	heta = \frac{1}{x}$,which implies $\cos 	heta = x$. Since $	an^2 	heta = \sec^2 	heta - 1$,we have $	an 	heta = \sqrt{(\frac{1}{x})^2 - 1} = \frac{\sqrt{1-x^2}}{x}$.Next,simplify the right side: Let $\phi = 	an^{-1}(2)$,then $	an \phi = 2$. Using the identity $\sin(	an^{-1} y) = \frac{y}{\sqrt{1+y^2}}$,we get $\sin \phi = \frac{2}{\sqrt{1+2^2}} = \frac{2}{\sqrt{5}}$.Equating both sides: $\frac{\sqrt{1-x^2}}{x} = \frac{2}{\sqrt{5}}$.Squaring both sides: $\frac{1-x^2}{x^2} = \frac{4}{5}$.$5(1-x^2) = 4x^2 \Rightarrow 5 - 5x^2 = 4x^2 \Rightarrow 9x^2 = 5 \Rightarrow x^2 = \frac{5}{9}$.Since $x>0$,we have $x = \frac{\sqrt{5}}{3}$.

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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