If sin^{-1} frac{1}{2} = tan^{-1} x, then x = | vedclass

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(B) Given the equation: $\sin^{-1} \frac{1}{2} = 	an^{-1} x$We know that $\sin^{-1} \frac{1}{2} = \frac{\pi}{6}$ because $\sin \frac{\pi}{6} = \frac{

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$\frac{1}{\sqrt{3}}$

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(B) Given the equation: $\sin^{-1} \frac{1}{2} = 	an^{-1} x$We know that $\sin^{-1} \frac{1}{2} = \frac{\pi}{6}$ because $\sin \frac{\pi}{6} = \frac{1}{2}$.Substituting this into the equation, we get: $\frac{\pi}{6} = 	an^{-1} x$Taking the tangent of both sides: $x = 	an \frac{\pi}{6}$Since $	an \frac{\pi}{6} = \frac{1}{\sqrt{3}}$, we have $x = \frac{1}{\sqrt{3}}$.

If $\sin^{-1} \frac{1}{2} = \tan^{-1} x$, then $x =$

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