If $\frac{1- an heta}{1+ an heta}=\frac{1}{\sqrt{3}}$, where $ heta \in\left(0, \frac{\pi}{2} ight)$, then $ heta=$

Question

If $\frac{1-	an 	heta}{1+	an 	heta}=\frac{1}{\sqrt{3}}$, where $	heta \in\left(0, \frac{\pi}{2}ight)$, then $	heta=$

Aicharya · Accepted Answer

Given $\frac{1-	an 	heta}{1+	an 	heta}=\frac{1}{\sqrt{3}}$ $	herefore 	an \left(\frac{\pi}{4}-	hetaight)=\frac{1}{\sqrt{3}}$ Comparing with $	an \frac{\pi}{6}=\frac{1}{\sqrt{3}}$, we write $\begin{aligned} & \frac{\pi}{4}-	heta=\frac{\pi}{6} \ 	herefore & 	heta=\frac{\pi}{4}-\frac{\pi}{6}=\frac{\pi}{12} \end{aligned}$

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

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