If $\frac{\cot \theta-1}{\cot \theta+1}=\frac{1-\sqrt{3}}{1+\sqrt{3}}$ then find the acute angle $\theta$.
Given : $\frac{\cot \theta-1}{\cot \theta+1}=\frac{1-\sqrt{3}}{1+\sqrt{3}}$
$\frac{\cot \theta-1}{\cot \theta+1}=\frac{1-\sqrt{3}}{1+\sqrt{3}}$
$\Rightarrow \frac{\frac{1}{\tan \theta}-1}{\frac{1}{\tan \theta}+1}=\frac{1-\sqrt{3}}{1+\sqrt{3}} \quad\left(\because \cot \theta=\frac{1}{\tan \theta}\right)$
$\Rightarrow \frac{\frac{1-\tan \theta}{\tan \theta}}{\frac{1+\tan \theta}{\tan \theta}}=\frac{1-\sqrt{3}}{1+\sqrt{3}}$
$\Rightarrow \frac{1-\tan \theta}{1+\tan \theta}=\frac{1-\sqrt{3}}{1+\sqrt{3}}$
On comparing LHS and RHS, we get
$\tan \theta=\sqrt{3}$
$\Rightarrow \theta=60^{\circ} \quad\left(\because \tan 60^{\circ}=\sqrt{3}\right)$
Hence, the acute angle $\theta$ is equal to $60^{\circ} .$