Find an acute angle $\theta$, when $\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }=\frac{1-\sqrt{3}}{1+\sqrt{3}}$.

$\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }=\frac{1-\sqrt{3}}{1+\sqrt{3}}$

$\frac{\frac{\cos \theta -\sin \theta }{\cos \theta }}{\frac{\cos \theta +\sin \theta }{\cos \theta }}=\frac{1-\sqrt{3}}{1+\sqrt{3}}$

$\frac{1-\tan \theta }{1+\tan \theta }=\frac{1-\sqrt{3}}{1+\sqrt{3}}$

On comparing the two sides, we get,

$\tan \theta =\sqrt{3}=\tan {60}^o$

$\theta ={60}^o$