1. Prove that \( \frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}=\sin \theta+\cos \theta \)

To prove the identity \[ \frac{\cos \theta}{1 - \tan \theta} + \frac{\sin \theta}{1 - \cot \theta} = \sin \theta + \cos \theta, \] we start by rewriting the left-hand side using the definitions of \(\tan\) and \(\cot\): \[ \tan \theta = \frac{\sin \theta}{\cos \theta}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta}. \] Thus, we can express \(1 - \tan \theta\) and \(1 - \cot \theta\) as follows: \[ 1 - \tan \theta = 1 - \frac{\sin \theta}{\cos \theta} = \frac{\cos \theta - \sin \theta}{\cos \theta}, \] \[ 1 - \cot \theta = 1 - \frac{\cos \theta}{\sin \theta} = \frac{\sin \theta - \cos \theta}{\sin \theta}. \] Now substituting these into the left-hand side: \[ \frac{\cos \theta}{1 - \tan \theta} = \frac{\cos \theta}{\frac{\cos \theta - \sin \theta}{\cos \theta}} = \frac{\cos^2 \theta}{\cos \theta - \sin \theta}, \] \[ \frac{\sin \theta}{1 - \cot \theta} = \frac{\sin \theta}{\frac{\sin \theta - \cos \theta}{\sin \theta}} = \frac{\sin^2 \theta}{\sin \theta - \cos \theta}. \] Now we can rewrite the left-hand side: \[ \frac{\cos^2 \theta}{\cos \theta - \sin \theta} + \frac{\sin^2 \theta}{\sin \theta - \cos \theta}. \] Notice that \(\sin \theta - \cos \theta = -(\cos \theta - \sin \theta)\), so we can rewrite the second term: \[ \frac{\sin^2 \theta}{\sin \theta - \cos \theta} = -\frac{\sin^2 \theta}{\cos \theta - \sin \theta}. \] Thus, we have: \[ \frac{\cos^2 \theta}{\cos \theta - \sin \theta} - \frac{\sin^2 \theta}{\cos \theta - \sin \theta} = \frac{\cos^2 \theta - \sin^2 \theta}{\cos \theta - \sin \theta}. \] Now we can factor the numerator: \[ \cos^2 \theta - \sin^2 \theta = (\cos \theta - \sin \theta)(\cos \theta + \sin \theta). \] Substituting this back into our expression gives: \[ \frac{(\cos \theta - \sin \theta)(\cos \theta + \sin \theta)}{\cos \theta - \sin \theta}. \] Assuming \(\cos \theta \neq \sin \theta\), we can cancel \(\cos \theta - \sin \theta\): \[ \cos \theta + \sin \theta. \] Thus, we have shown that: \[ \frac{\cos \theta}{1 - \tan \theta} + \frac{\sin \theta}{1 - \cot \theta} = \sin \theta + \cos \theta. \] Therefore, the identity is proven: \[ \boxed{\sin \theta + \cos \theta}. \]