LET US PROVE Prove the following identities when they are defined: 1. \( \cos ^{2} x-\cos ^{2} y=\sin ^{2} y-\sin ^{2} \chi \) 2. \( \sin ^{4} x-\cos ^{4} x=(\sin x-\cos x)(\sin x+\cos x) \) 3. \( \frac{1-\cos x}{\sin x}-\frac{\sin x}{1+\cos x}=0 \) (4.) \( \frac{1}{1+\tan ^{2} x}-\frac{1}{1+\tan ^{2} y}=\sin ^{2} y-\sin ^{2} x \)

Let's prove each of the identities one by one. ### Identity 1: \[ \cos^2 x - \cos^2 y = \sin^2 y - \sin^2 x \] **Proof:** We can use the Pythagorean identity, which states that \(\sin^2 z + \cos^2 z = 1\). Starting with the left-hand side: \[ \cos^2 x - \cos^2 y = (\cos^2 x - 1) - (\cos^2 y - 1) = -\sin^2 x + \sin^2 y \] Thus, we have: \[ \cos^2 x - \cos^2 y = \sin^2 y - \sin^2 x \] This proves the identity. ### Identity 2: \[ \sin^4 x - \cos^4 x = (\sin x - \cos x)(\sin x + \cos x) \] **Proof:** We can use the difference of squares on the left-hand side: \[ \sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x) \] Since \(\sin^2 x + \cos^2 x = 1\), we have: \[ \sin^4 x - \cos^4 x = \sin^2 x - \cos^2 x \] Now, we can factor \(\sin^2 x - \cos^2 x\) as: \[ \sin^2 x - \cos^2 x = (\sin x - \cos x)(\sin x + \cos x) \] Thus, we conclude: \[ \sin^4 x - \cos^4 x = (\sin x - \cos x)(\sin x + \cos x) \] This proves the identity. ### Identity 3: \[ \frac{1 - \cos x}{\sin x} - \frac{\sin x}{1 + \cos x} = 0 \] **Proof:** We will find a common denominator for the left-hand side: The common denominator is \(\sin x(1 + \cos x)\). Thus, we rewrite the left-hand side: \[ \frac{(1 - \cos x)(1 + \cos x) - \sin^2 x}{\sin x(1 + \cos x)} \] Now, simplify the numerator: \[ (1 - \cos x)(1 + \cos x) = 1 - \cos^2 x = \sin^2 x \] So, we have: \[ \sin^2 x - \sin^2 x = 0 \] Thus, the left-hand side simplifies to: \[ \frac{0}{\sin x(1 + \cos x)} = 0 \] This proves the identity. ### Identity 4: \[ \frac{1}{1 + \tan^2 x} - \frac{1}{1 + \tan^2 y} = \sin^2 y - \sin^2 x \] **Proof:** Recall that \(1 + \tan^2 z = \frac{1}{\cos^2 z}\). Therefore: \[ \frac{1}{1 + \tan^2 x} = \cos^2 x \quad \text{and} \quad \frac{1}{1 + \tan^2 y} = \cos^2 y \] Thus, the left-hand side becomes: \[ \cos^2 x - \cos^2 y \] Using the identity \(\sin^2 z = 1 - \cos^2 z\), we can rewrite the right-hand side: \[ \sin^2 y - \sin^2 x = (1 - \cos^2 y) - (1 - \cos^2 x) = \cos^2 x - \cos^2 y \] Thus, we have: \[ \cos^2 x - \cos^2 y = \sin^2 y - \sin^2 x \] This proves the identity. In conclusion, all four identities have been proven.