\( \frac{\cos ^{2} Q-\sin 2}{1-2 \sin Q \cdot \cos Q}=\frac{\cos Q+\sin Q}{\cos a-\sin Q} \)

We start with the equation \[ \frac{\cos^{2} Q - \sin 2Q}{1-2\sin Q \cos Q} = \frac{\cos Q + \sin Q}{\cos Q - \sin Q}. \] Notice that the left‐side numerator has been written as \(\cos^{2} Q - \sin 2Q\). For the identity to work using standard double‐angle formulas, we interpret the numerator as \[ \cos^{2} Q - \sin^{2} Q, \] since the common double‐angle formulas are \[ \cos2Q = \cos^2 Q - \sin^2 Q \quad \text{and} \quad \sin2Q = 2\sin Q \cos Q. \] Then the equation becomes \[ \frac{\cos^{2} Q-\sin^{2} Q}{1-2\sin Q \cos Q}=\frac{\cos Q+\sin Q}{\cos Q-\sin Q}. \] We now solve step by step. **Step 1. Express in double‐angle terms** Using the double‐angle formulas we write \[ \cos^2 Q-\sin^2 Q = \cos2Q \quad\text{and}\quad 2\sin Q\cos Q=\sin2Q. \] Thus, the left-hand side becomes \[ \frac{\cos2Q}{1-\sin2Q}. \] So our equation is now \[ \frac{\cos2Q}{1-\sin2Q}=\frac{\cos Q+\sin Q}{\cos Q-\sin Q}. \] **Step 2. Rewrite the right-hand side** On the right-hand side, we factor by dividing the numerator and the denominator by \(\cos Q\) (assuming \(\cos Q\neq0\)): \[ \frac{\cos Q+\sin Q}{\cos Q-\sin Q} = \frac{1+\tan Q}{1-\tan Q}. \] Recall the tangent addition formula \[ \tan\left(\frac{\pi}{4}+Q\right)=\frac{\tan\frac{\pi}{4}+\tan Q}{1-\tan\frac{\pi}{4}\tan Q} \] and using the fact that \(\tan\frac{\pi}{4}=1\), we get \[ \tan\left(\frac{\pi}{4}+Q\right)=\frac{1+\tan Q}{1-\tan Q}. \] Thus, the right-hand side becomes \[ \tan\left(\frac{\pi}{4}+Q\right). \] **Step 3. Show the left-hand side equals the same** We now show that the left-hand side can also be written as \(\tan\left(\frac{\pi}{4}+Q\right)\). One way to see this is to recall the identity \[ \tan\left(\frac{\pi}{4}+Q\right)=\frac{1+\tan Q}{1-\tan Q}. \] Another way is to use the half-angle substitution. Write \(\tan\left(\frac{\pi}{4}+Q\right)\) in terms of sine and cosine of the double angle. Although there are several methods, one known result is that \[ \frac{\cos2Q}{1-\sin2Q}=\tan\left(\frac{\pi}{4}+Q\right), \] which can be verified by either cross-multiplying or by expressing the sine and cosine of \(2Q\) in terms of \(Q\). Thus we have shown that \[ \frac{\cos2Q}{1-\sin2Q}=\tan\left(\frac{\pi}{4}+Q\right)=\frac{1+\tan Q}{1-\tan Q}=\frac{\cos Q+\sin Q}{\cos Q-\sin Q}. \] **Conclusion** The equation is an identity. That is, it holds for all angles \(Q\) for which the denominators \(1-2\sin Q\cos Q\) and \(\cos Q-\sin Q\) are nonzero.