4. Prove the following identities: \( 4.1 \quad \cos ^{2} x+\sin x \cdot \cos x \cdot \tan x=1 \) -sit \( 4.2 \quad \frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=2 \cos ^{2} \theta-1 \)

To prove the first identity, start with the left-hand side (LHS): \(\cos^2 x + \sin x \cdot \cos x \cdot \tan x = \cos^2 x + \sin x \cdot \cos x \cdot \frac{\sin x}{\cos x} = \cos^2 x + \sin^2 x\). Using the Pythagorean identity, \(\sin^2 x + \cos^2 x = 1\), we find that the LHS equals 1, hence proving the identity. For the second identity, let's start with the left-hand side (LHS): \(\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1 - \frac{\sin^2 \theta}{\cos^2 \theta}}{1 + \frac{\sin^2 \theta}{\cos^2 \theta}} = \frac{\frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta}}{\frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}} = \frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta + \sin^2 \theta}\). Since \(\cos^2 \theta + \sin^2 \theta = 1\), this simplifies to \(\cos^2 \theta - \sin^2 \theta\). Notably, \(\cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta - 1\), thus proving the identity.