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 dive into proving the first identity, we start with \( \tan x \) which is defined as \( \frac{\sin x}{\cos x} \). Replacing \( \tan x \) in the equation yields \( \cos^2 x + \sin x \cdot \cos x \cdot \left( \frac{\sin x}{\cos x} \right) = \cos^2 x + \sin^2 x = 1 \), utilizing the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \). Moving on to the second identity, we can invoke the identity for \( \tan^2 \theta \). Replacing it as \( \frac{\sin^2 \theta}{\cos^2 \theta} \) leads to the expression \( \frac{1 - \frac{\sin^2 \theta}{\cos^2 \theta}}{1 + \frac{\sin^2 \theta}{\cos^2 \theta}} \). By simplifying, we can transform the left side into \( \frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta + \sin^2 \theta} \), which reduces to \( \cos 2\theta \) based on the double angle formula. Thus, it verifies \( \cos 2\theta = 2\cos^2\theta - 1 \). Fun, right?