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

1. Start with the given expression: \[ \frac{1-\cos x}{\sin x} - \frac{\sin x}{1+\cos x} \] 2. Consider rewriting the second fraction by rationalizing its denominator. Multiply the numerator and denominator of the second term by \((1-\cos x)\): \[ \frac{\sin x}{1+\cos x} = \frac{\sin x(1-\cos x)}{(1+\cos x)(1-\cos x)} \] 3. Notice that the denominator simplifies using the difference of squares: \[ (1+\cos x)(1-\cos x) = 1 - \cos^2 x \] 4. Recall the Pythagorean identity: \[ 1 - \cos^2 x = \sin^2 x \] Therefore, \[ \frac{\sin x(1-\cos x)}{(1+\cos x)(1-\cos x)} = \frac{\sin x(1-\cos x)}{\sin^2 x} \] 5. Simplify the fraction by canceling a common factor of \(\sin x\) in the numerator and denominator: \[ \frac{\sin x(1-\cos x)}{\sin^2 x} = \frac{1-\cos x}{\sin x} \] 6. Substitute this back into the original expression: \[ \frac{1-\cos x}{\sin x} - \frac{1-\cos x}{\sin x} = 0 \] 7. Thus, the identity is proven: \[ \frac{1-\cos x}{\sin x} - \frac{\sin x}{1+\cos x} = 0 \]