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 tackle the first identity \( \cos^{2} x + \sin x \cdot \cos x \cdot \tan x = 1 \), let's rewrite \( \tan x \) as \( \frac{\sin x}{\cos x} \). Substituting this into the equation gives: \[ \cos^{2} x + \sin x \cdot \cos x \cdot \left(\frac{\sin x}{\cos x}\right) = \cos^{2} x + \sin^{2} x = 1 \] This uses the Pythagorean identity \( \sin^{2} x + \cos^{2} x = 1 \). Now for the identity \( \frac{1-\tan^{2} \theta}{1+\tan^{2} \theta} = 2 \cos^{2} \theta - 1 \), we start by expressing \( \tan^{2} \theta \) as \( \frac{\sin^{2} \theta}{\cos^{2} \theta} \): \[ \frac{1 - \frac{\sin^{2} \theta}{\cos^{2} \theta}}{1 + \frac{\sin^{2} \theta}{\cos^{2} \theta}} = \frac{\cos^{2} \theta - \sin^{2} \theta}{\cos^{2} \theta + \sin^{2} \theta} = \cos^{2} \theta - \sin^{2} \theta \] Since \( \cos^{2} \theta + \sin^{2} \theta = 1 \), we can simplify further to show this equals \( 2 \cos^{2} \theta - 1 \). Both identities hold true!