Question

Simplify:
$\sqrt{\frac{1-\sin 2\theta }{1+\sin 2\theta }}$

Answer 1

Now,

$\sqrt{\frac{1-\sin 2\theta }{1+\sin 2\theta }}$

$=\sqrt{\frac{{\sin }^2\theta +{\cos }^2\theta -2\sin \theta .\cos \theta }{{\sin }^2\theta +{\cos }^2\theta +2\sin \theta .\cos \theta }}$

$=\sqrt{\frac{(\sin \theta -\cos \theta {)}^2}{(\sin \theta +\cos \theta {)}^2}}$

$=\frac{(\sin \theta -\cos \theta )}{(\sin \theta +\cos \theta )}$

$=\frac{(\tan \theta -1)}{(\tan \theta +1)}$ [ Dividing the numerator and the denominator by $\cos \theta$]

$=\frac{(\tan \theta -\tan \frac{\pi }{4})}{(1+\tan \theta .\tan \frac{\pi }{4})}$

$=\tan (\theta -\frac{\pi }{4})$