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

Let's prove both identities one by one! **a)** We start with the left side: \[ \sin x \cdot \tan x + \cos x \] Recall that \(\tan x = \frac{\sin x}{\cos x}\), so we can substitute: \[ \sin x \cdot \frac{\sin x}{\cos x} + \cos x = \frac{\sin^2 x}{\cos x} + \cos x \] To combine these terms, we rewrite \(\cos x\) as \(\frac{\cos^2 x}{\cos x}\): \[ \frac{\sin^2 x}{\cos x} + \frac{\cos^2 x}{\cos x} = \frac{\sin^2 x + \cos^2 x}{\cos x} \] Using the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\): \[ \frac{1}{\cos x} \] Thus, we have established that: \[ \sin x \cdot \tan x + \cos x = \frac{1}{\cos x} \] **b)** We start with the left side: \[ \frac{\cos x}{1-\sin x} - \frac{\cos x}{1+\sin x} \] To combine these fractions, we need a common denominator, which is \((1 - \sin x)(1 + \sin x)\). Therefore, we rewrite the fractions: \[ \frac{\cos x(1 + \sin x) - \cos x(1 - \sin x)}{(1 - \sin x)(1 + \sin x)} \] This simplifies to: \[ \frac{\cos x + \cos x \sin x - \cos x + \cos x \sin x}{(1 - \sin x)(1 + \sin x)} \] Notice that \(\cos x\) cancels, so we have: \[ \frac{2\cos x \sin x}{1 - \sin^2 x} \] Using the identity \(1 - \sin^2 x = \cos^2 x\): \[ \frac{2\cos x \sin x}{\cos^2 x} = 2 \left(\frac{\sin x}{\cos x}\right) = 2\tan x \] Therefore, we have shown that: \[ \frac{\cos x}{1-\sin x}-\frac{\cos x}{1+\sin x} = 2 \tan x \]