Question

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

Answer 1

Now,

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

$=\sqrt{\frac{{\sin }^{2}x+{\cos }^{2}x+2\sin x.\cos x}{{\sin }^{2}x+{\cos }^{2}x-2\sin x.\cos x}}$ [ Since ${\sin }^{2}x+{\cos }^{2}x=1$]

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

$=\frac{\sin x+\cos x}{\cos x-\sin x}$

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

$=\frac{\tan \frac{\pi }{4}+\tan x}{1-\tan x.\tan \frac{\pi }{4}}$

$=\tan (\frac{\pi }{4}+x)$.