If {sin ^{ - 1}}frac{1}{2} = {tan ^{ - 1}}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}

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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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