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, we start with the left side: \[ \cos^2 x + \sin x \cdot \cos x \cdot \tan x \] Recall that \(\tan x = \frac{\sin x}{\cos x}\), so we can rewrite the term \(\sin x \cdot \cos x \cdot \tan x\) as: \[ \sin x \cdot \cos x \cdot \frac{\sin x}{\cos x} = \sin^2 x \] Thus, the left side becomes: \[ \cos^2 x + \sin^2 x \] Using the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\), we confirm that: \[ \cos^2 x + \sin^2 x = 1 \] So, the identity is proven. For the second identity, we start with the left side: \[ \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \] Using the definition of \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\), we can rewrite this as: \[ \frac{1 - \frac{\sin^2 \theta}{\cos^2 \theta}}{1 + \frac{\sin^2 \theta}{\cos^2 \theta}} = \frac{\frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta}}{\frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}} = \frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta + \sin^2 \theta} \] Since \(\cos^2 \theta + \sin^2 \theta = 1\), this simplifies to: \[ \cos^2 \theta - \sin^2 \theta \] Now, using another trigonometric identity, we know: \[ \cos(2\theta) = \cos^2 \theta - \sin^2 \theta \] Thus, we have: \[ \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos(2\theta) \] And since \(2 \cos^2 \theta - 1 = \cos(2\theta)\), the right side also equals \(\cos(2\theta)\). Hence, the identity is confirmed!