1. Prove the folloking a) \( \frac{\sin \theta-\tan \theta \cos ^{2} \theta}{\cos \theta-1+\sin \theta} \)

We start with the given expression \[ \frac{\sin \theta - \tan \theta\, \cos^2 \theta}{\cos \theta-1+\sin \theta}. \] Our goal is to simplify it step by step. **Step 1. Rewrite \(\tan\theta\) in terms of \(\sin\theta\) and \(\cos\theta\).** Recall that \[ \tan \theta=\frac{\sin\theta}{\cos\theta}. \] Substitute this into the numerator: \[ \sin \theta - \tan \theta\, \cos^2\theta = \sin \theta - \frac{\sin\theta}{\cos\theta}\,\cos^2\theta. \] **Step 2. Simplify the numerator.** Notice that \[ \frac{\sin\theta}{\cos\theta}\,\cos^2\theta = \sin\theta\,\cos\theta. \] Thus the numerator becomes \[ \sin \theta - \sin\theta\,\cos\theta = \sin\theta\,(1-\cos\theta). \] **Step 3. Write the expression with the simplified numerator.** The given expression now is \[ \frac{\sin\theta\,(1-\cos\theta)}{\cos\theta-1+\sin\theta}. \] It is often useful to rewrite the denominator in a form that makes any common structure more visible. Notice that \[ \cos\theta-1+\sin\theta=\sin\theta+\cos\theta-1. \] So we can write the expression as \[ \frac{\sin\theta\,(1-\cos\theta)}{\sin\theta+\cos\theta-1}. \] **Step 4. Comment on the result.** At this stage the expression is simplified to \[ \frac{\sin\theta\,(1-\cos\theta)}{\sin\theta+\cos\theta-1}. \] No further common factors appear between the numerator and the denominator. Hence, we have proved that \[ \frac{\sin \theta - \tan \theta\, \cos^2\theta}{\cos \theta-1+\sin \theta} =\frac{\sin\theta\,(1-\cos\theta)}{\sin\theta+\cos\theta-1}. \] This completes the simplification (proof) of the given expression.