Prove that: a) \( \sin x \cdot \tan x+\cos x=\frac{1}{\cos x} \) b) \( \frac{\cos x}{1-\sin x}-\frac{\cos x}{1+\sin x}=2 \tan x \)

To prove each of these trigonometric identities, let's tackle them one by one! a) Starting with the left-hand side: \[ \sin x \cdot \tan x + \cos x \] We know that \(\tan x = \frac{\sin x}{\cos x}\), so we substitute: \[ \sin x \cdot \frac{\sin x}{\cos x} + \cos x = \frac{\sin^2 x}{\cos x} + \cos x \] Now, we can rewrite \(\cos x\) with a common denominator: \[ \frac{\sin^2 x}{\cos x} + \frac{\cos^2 x}{\cos x} = \frac{\sin^2 x + \cos^2 x}{\cos x} \] Using the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\): \[ \frac{1}{\cos x} \] Thus, we have shown: \[ \sin x \cdot \tan x + \cos x = \frac{1}{\cos x} \] b) Now for the second part: Starting with the left-hand side: \[ \frac{\cos x}{1 - \sin x} - \frac{\cos x}{1 + \sin x} \] We can find a common denominator, which is \((1 - \sin x)(1 + \sin x) = 1 - \sin^2 x\). The expression simplifies to: \[ \frac{\cos x(1 + \sin x) - \cos x(1 - \sin x)}{(1 - \sin x)(1 + \sin x)} \] This simplifies further: \[ \frac{\cos x + \cos x \sin x - \cos x + \cos x \sin x}{1 - \sin^2 x} = \frac{2 \cos x \sin x}{1 - \sin^2 x} \] Since \(1 - \sin^2 x = \cos^2 x\), we have: \[ \frac{2 \cos x \sin x}{\cos^2 x} = 2 \cdot \frac{\sin x}{\cos x} = 2 \tan x \] Thus, we have proven: \[ \frac{\cos x}{1 - \sin x} - \frac{\cos x}{1 + \sin x} = 2 \tan x \] Both identities are proven!