4.4 Prove that the identity: \[ \frac{\cos ^{2} A-\sin ^{2} A}{1+2 \sin A \cdot \sin \left(90^{\circ}-A\right)}=\frac{\cos A-\sin A}{\sin A+\cos A} \]

To prove the identity \[ \frac{\cos^{2} A - \sin^{2} A}{1 + 2 \sin A \cdot \sin(90^{\circ} - A)} = \frac{\cos A - \sin A}{\sin A + \cos A}, \] we start with the left-hand side (LHS): 1. We know that \(\sin(90^{\circ} - A) = \cos A\), so we can simplify the LHS: \[ \text{LHS} = \frac{\cos^{2} A - \sin^{2} A}{1 + 2 \sin A \cdot \cos A}. \] 2. Use the identity \(\cos^{2} A - \sin^{2} A = \cos(2A)\) and \(1 + 2 \sin A \cos A = 1 + \sin(2A)\): \[ \text{LHS} = \frac{\cos(2A)}{1 + \sin(2A)}. \] Now let’s simplify the right-hand side (RHS): 3. The RHS can be rewritten as: \[ \text{RHS} = \frac{\cos A - \sin A}{\sin A + \cos A}. \] 4. To relate both sides, multiply the numerator and denominator of the RHS by \((\cos A + \sin A)\): \[ \text{RHS} = \frac{(\cos A - \sin A)(\cos A + \sin A)}{(\sin A + \cos A)(\cos A + \sin A)} = \frac{\cos^{2} A - \sin^{2} A}{\sin^{2} A + 2 \sin A \cos A + \cos^{2} A}. \] 5. Notice that \(\sin^{2} A + \cos^{2} A = 1\), thus the denominator simplifies to \(1 + 2 \sin A \cos A\): \[ \text{RHS} = \frac{\cos^{2} A - \sin^{2} A}{1 + 2 \sin A \cos A}. \] 6. Now we can see that: \[ \text{RHS} = \text{LHS}. \] Thus, we have shown that \[ \frac{\cos^{2} A - \sin^{2} A}{1 + 2 \sin A \cdot \sin(90^{\circ} - A)} = \frac{\cos A - \sin A}{\sin A + \cos A}. \] This proves the identity.