Prove that \( \frac{\operatorname{Sin} \theta(1+\operatorname{Cos} \theta)}{1-\cos ^{2} \theta}+\tan \theta=\frac{1+\operatorname{Cos}}{\sin \theta \cdot Q} \)

To prove the equation \[ \frac{\sin \theta(1+\cos \theta)}{1-\cos^2 \theta} + \tan \theta = \frac{1+\cos \theta}{\sin \theta \cdot Q}, \] we will simplify the left-hand side and see if it can be transformed into the right-hand side. ### Step 1: Simplify the left-hand side The left-hand side is \[ \frac{\sin \theta(1+\cos \theta)}{1-\cos^2 \theta} + \tan \theta. \] We know that \[ 1 - \cos^2 \theta = \sin^2 \theta. \] Thus, we can rewrite the first term: \[ \frac{\sin \theta(1+\cos \theta)}{\sin^2 \theta} = \frac{1+\cos \theta}{\sin \theta}. \] Now, substituting this back into the left-hand side gives us: \[ \frac{1+\cos \theta}{\sin \theta} + \tan \theta. \] ### Step 2: Rewrite \(\tan \theta\) Recall that \[ \tan \theta = \frac{\sin \theta}{\cos \theta}. \] So, we can rewrite the left-hand side as: \[ \frac{1+\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta}. \] ### Step 3: Find a common denominator The common denominator for the two fractions is \(\sin \theta \cos \theta\). Thus, we can rewrite the left-hand side as: \[ \frac{(1+\cos \theta) \cos \theta + \sin^2 \theta}{\sin \theta \cos \theta}. \] ### Step 4: Simplify the numerator Now, let's simplify the numerator: \[ (1+\cos \theta) \cos \theta + \sin^2 \theta = \cos \theta + \cos^2 \theta + \sin^2 \theta. \] Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\), we can simplify this to: \[ \cos \theta + 1. \] ### Step 5: Substitute back into the left-hand side Now, substituting this back, we have: \[ \frac{\cos \theta + 1}{\sin \theta \cos \theta}. \] ### Step 6: Compare with the right-hand side Now, we need to compare this with the right-hand side: \[ \frac{1+\cos \theta}{\sin \theta \cdot Q}. \] From our simplification, we see that: \[ \frac{\cos \theta + 1}{\sin \theta \cos \theta} = \frac{1+\cos \theta}{\sin \theta \cdot Q}. \] This implies that: \[ Q = \cos \theta. \] ### Conclusion Thus, we have shown that: \[ \frac{\sin \theta(1+\cos \theta)}{1-\cos^2 \theta} + \tan \theta = \frac{1+\cos \theta}{\sin \theta \cdot \cos \theta}. \] Therefore, the original equation holds true, and we have proven the statement.