5.4 Consider the identity: \( \frac{1-\sin 2 \mathrm{~A}}{\sin \mathrm{~A}-\cos \mathrm{A}}=\sin \mathrm{A}-\cos \mathrm{A} \) 5.4.1 Prove the identity. 5.4.2 For which values of A in the interval \( 0^{\circ}<\mathrm{A}<180^{\circ} \) will the identity be

To prove the identity \[ \frac{1 - \sin 2A}{\sin A - \cos A} = \sin A - \cos A, \] we will start by simplifying the left-hand side. ### Step 1: Simplify the left-hand side Recall that \(\sin 2A = 2 \sin A \cos A\). Therefore, we can rewrite the left-hand side: \[ 1 - \sin 2A = 1 - 2 \sin A \cos A. \] Now, substituting this into the left-hand side gives: \[ \frac{1 - 2 \sin A \cos A}{\sin A - \cos A}. \] ### Step 2: Factor the numerator We can factor the numerator \(1 - 2 \sin A \cos A\) using the identity \(1 - 2xy = (1 - x)^2 + (1 - y)^2\) or recognize it as a difference of squares. However, a more straightforward approach is to rewrite it as: \[ 1 - 2 \sin A \cos A = (1 - \sin A)(1 + \sin A) - 2 \sin A \cos A. \] This doesn't simplify directly, so let's proceed with the division. ### Step 3: Cross-multiply to prove the identity To prove the identity, we can cross-multiply: \[ 1 - 2 \sin A \cos A = (\sin A - \cos A)(\sin A - \cos A). \] Expanding the right-hand side: \[ (\sin A - \cos A)^2 = \sin^2 A - 2 \sin A \cos A + \cos^2 A. \] Using the Pythagorean identity \(\sin^2 A + \cos^2 A = 1\), we can rewrite this as: \[ 1 - 2 \sin A \cos A. \] ### Step 4: Equate both sides Now we have: \[ 1 - 2 \sin A \cos A = 1 - 2 \sin A \cos A. \] This confirms that both sides are equal, thus proving the identity. ### Step 5: Determine the values of A Now, we need to find the values of \(A\) in the interval \(0^\circ < A < 180^\circ\) for which the identity holds. The identity is valid as long as the denominator \(\sin A - \cos A\) is not equal to zero. Therefore, we need to solve: \[ \sin A - \cos A = 0. \] This implies: \[ \sin A = \cos A. \] The solutions to this equation occur at: \[ A = 45^\circ + n \cdot 180^\circ, \] where \(n\) is an integer. In the interval \(0^\circ < A < 180^\circ\), the only solution is: \[ A = 45^\circ. \] ### Conclusion Thus, the identity holds for all values of \(A\) in the interval \(0^\circ < A < 180^\circ\) except at \(A = 45^\circ\). In summary: - **5.4.1**: The identity is proven. - **5.4.2**: The identity holds for \(0^\circ < A < 180^\circ\) except at \(A = 45^\circ\).