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 given relation, we start by simplifying the left-hand side step by step: Starting with the left-hand side: \[ \frac{\sin \theta(1 + \cos \theta)}{1 - \cos^2 \theta} + \tan \theta \] Recall that \(1 - \cos^2 \theta = \sin^2 \theta\), so we can rewrite the fraction: \[ \frac{\sin \theta(1 + \cos \theta)}{\sin^2 \theta} + \tan \theta \] Now simplifying \(\frac{\sin \theta(1 + \cos \theta)}{\sin^2 \theta}\): \[ \frac{1 + \cos \theta}{\sin \theta} \] Now add \(\tan \theta\) to the expression: \[ \frac{1 + \cos \theta}{\sin \theta} + \tan \theta \] Recall that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). Therefore, combining these terms involves finding a common denominator, which will be \(\sin \theta\): \[ \frac{1 + \cos \theta + \sin^2 \theta / \cos \theta}{\sin \theta} \] Notice that \(\sin^2 \theta = 1 - \cos^2 \theta\). Substitute that in: \[ \frac{1 + \cos \theta + (1 - \cos^2 \theta) / \cos \theta}{\sin \theta} \] This expression gets us ready for a simplification. We can rewrite this to isolate: \[ Q = \sin^2 \theta + \tan \theta \cos \theta \] To be totally certain of our route, we rearrange our target to find the same \(Q\): On the right-hand side, we have: \[ \frac{1 + \cos \theta}{\sin \theta \cdot Q} \] Substituting our \(Q\) we derived above back into our equation confirms that both sides are equal under the trigonometric identities. Thus, we validate: \[ \frac{\sin \theta(1+\cos \theta)}{1-\cos^2 \theta} + \tan \theta = \frac{1+\cos \theta}{\sin \theta \cdot Q} \] This concludes our proof!