Prove chat \( \frac{\sin \theta}{1+\cos \theta}=\frac{1-\cos \theta}{\sin \theta} \)

To prove the identity \( \frac{\sin \theta}{1+\cos \theta} = \frac{1-\cos \theta}{\sin \theta} \), we can start by cross multiplying to eliminate the fractions: \[ \sin^2 \theta = (1 - \cos \theta)(1 + \cos \theta) \] The right-hand side simplifies using the difference of squares: \[ (1 - \cos \theta)(1 + \cos \theta) = 1 - \cos^2 \theta \] We know from the Pythagorean identity that \( \sin^2 \theta + \cos^2 \theta = 1 \), which means \( 1 - \cos^2 \theta = \sin^2 \theta \). Thus, we have: \[ \sin^2 \theta = \sin^2 \theta \] This confirms that the two sides are equal, proving the identity correct. So, the statement \( \frac{\sin \theta}{1+\cos \theta} = \frac{1-\cos \theta}{\sin \theta} \) holds true!