If frac{1-tan theta}{1+tan theta}=frac{1}{sqr | vedclass

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(A) Given the equation $\frac{1-	an 	heta}{1+	an 	heta}=\frac{1}{\sqrt{3}}$.Using the formula $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an

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$\frac{\pi}{12}$

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(A) Given the equation $\frac{1-	an 	heta}{1+	an 	heta}=\frac{1}{\sqrt{3}}$.Using the formula $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we know that $	an \left(\frac{\pi}{4}-	hetaight) = \frac{	an \frac{\pi}{4} - 	an 	heta}{1 + 	an \frac{\pi}{4} 	an 	heta} = \frac{1-	an 	heta}{1+	an 	heta}$.Therefore,$	an \left(\frac{\pi}{4}-	hetaight) = \frac{1}{\sqrt{3}}$.Since $	an \frac{\pi}{6} = \frac{1}{\sqrt{3}}$,we have $\frac{\pi}{4}-	heta = \frac{\pi}{6}$.Solving for $	heta$: $	heta = \frac{\pi}{4} - \frac{\pi}{6} = \frac{3\pi - 2\pi}{12} = \frac{\pi}{12}$.

If $\frac{1-\tan \theta}{1+\tan \theta}=\frac{1}{\sqrt{3}}$,where $\theta \in \left(0, \frac{\pi}{2}\right)$,then $\theta=$

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