It is given that, $$\sqrt{\frac{1 - \sin x}{1 + \sin x}} = a - \tan x$$ then a is equal to:

$$\sqrt{\frac{1 - \sin x}{1 + \sin x}} = a - \tan x$$

$$\sqrt{\frac{1 - \sin x}{1 + \sin x}\times\frac{1 - \sin x}{1 - \sin x}}=a - \tan x$$

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

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

$$\frac{1-\sin x}{\cos x}=a-\frac{\sin x}{\cos x}$$ , Because $$\sin^{2}x+\cos^{2}x=1$$

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

$$a=\frac{1}{\cos x}=\sec x$$