prove lenentify \( \frac{1-\cos x}{\sin x}-\frac{\sin x}{1+\cos x}=0 \)

Let's start by simplifying the given expression step-by-step. 1. We have: \[ \frac{1 - \cos x}{\sin x} - \frac{\sin x}{1 + \cos x} \] To combine these fractions, we need a common denominator. The common denominator will be \(\sin x (1 + \cos x)\). 2. Rewriting both fractions with the common denominator: \[ \frac{(1 - \cos x)(1 + \cos x)}{\sin x (1 + \cos x)} - \frac{\sin^2 x}{\sin x (1 + \cos x)} \] 3. Now, combine the numerators: \[ \frac{(1 - \cos x)(1 + \cos x) - \sin^2 x}{\sin x (1 + \cos x)} \] 4. Expanding the first part: \[ (1 - \cos^2 x) = \sin^2 x \] Thus, we can rewrite the numerator: \[ \sin^2 x - \sin^2 x = 0 \] 5. Therefore, we have: \[ \frac{0}{\sin x (1 + \cos x)} = 0 \] Thus, the original expression simplifies to zero, proving that: \[ \frac{1 - \cos x}{\sin x} - \frac{\sin x}{1 + \cos x} = 0 \] And that’s how we can prove the identity!