How to prove $\sqrt\frac{1 - \sin x}{1 + \sin x} = \frac{1}{\cos x} - \tan x$ Prove $\sqrt\frac{1 - \sin x}{1 + \sin x} = \frac{1}{\cos x} - \tan x$

I tried but I couldn't figure it out, give me a hint please.
Hint: Multiply the quantity under the root by $$\frac{1-\sin(x)}{1-\sin(x)}$$
$\sqrt\frac{1 - \sin x}{1 + \sin x} $

=$\sqrt\frac{(1 - \sin x)(1 - \sin x)}{(1 - \sin x)(1 + \sin x)}$

=$\sqrt\frac{(1 - \sin x)^2}{(1 - \sin^2 x)}$

=$\sqrt\frac{(1 - \sin x)^2}{(\cos^2x)}$           [because $sin^2 x + \cos^2 x =1$]

=$\frac{1-\sin x}{\cos x}$

=$\frac{1}{cosx}-\frac{\sin x}{\cos x}$

=$\frac{1}{cosx}-\tan x$

NOTE : This answer holds only if $sine$ and $cosine$ functions are in first quadrant.