\( \frac{\cos ^{2} Q-\sin 2}{1-2 \sin Q \cdot \cos Q}=\frac{\cos Q+\sin Q}{\cos a-\sin Q} \)
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To solve the equation \( \frac{\cos^{2} Q - \sin 2}{1 - 2 \sin Q \cdot \cos Q} = \frac{\cos Q + \sin Q}{\cos a - \sin Q} \), let's first simplify both sides. 1. Recall that \( \sin 2 = 2\sin Q \cos Q \). Thus, the left side becomes: \[ \frac{\cos^{2} Q - 2\sin Q \cos Q}{1 - 2\sin Q \cos Q} \] 2. Let's break down the numerator: - The numerator simplifies to \( \cos^{2} Q - 2\sin Q \cos Q = \cos^{2} Q - 2\sin Q \cos Q = (\cos Q - \sin Q)^{2} \). 3. Therefore, the left side can be rewritten as: \[ \frac{(\cos Q - \sin Q)^{2}}{1 - 2\sin Q \cos Q} \] Now we can refer to the right side, \( \frac{\cos Q + \sin Q}{\cos a - \sin Q} \). If you wish to find a common solution for angles \( Q \) and \( a \), it might be helpful to test specific values or manipulate the equation further. Checking if \( Q = a \) or other standard angles that relate cosine and sine could yield results, such as \( Q = 45^\circ \).