\( \cos (A-B) \) to prove that \( \cos (A-B)=\cos A \cdot \cos B+\sin A \cdot \sin B \) nplify: \( \frac{\cos \left(\theta-90^{\circ}\right)}{} \)

1. We start with the cosine subtraction formula: \[ \cos (A-B) = \cos A \cdot \cos B + \sin A \cdot \sin B. \] 2. To verify this identity, one approach is to derive it using the unit circle. - Consider two angles, \( A \) and \( B \), on the unit circle. - The coordinates corresponding to angle \( A \) are \( (\cos A, \sin A) \) and for angle \( B \) are \( (\cos B, \sin B) \). - Using the definition of the dot product between these unit vectors, and knowing that the dot product is equal to the cosine of the angle between the vectors, one can show that: \[ \cos (A-B) = \cos A \cdot \cos B + \sin A \cdot \sin B. \] 3. Next, let us simplify the expression: \[ \frac{\cos \left(\theta-90^{\circ}\right)}{} \] Although the fraction bar appears unnecessary here, the main expression to simplify is: \[ \cos \left(\theta-90^{\circ}\right). \] 4. We apply the cosine subtraction formula with \( A = \theta \) and \( B = 90^{\circ} \): \[ \cos \left(\theta-90^{\circ}\right) = \cos \theta \cdot \cos 90^{\circ} + \sin \theta \cdot \sin 90^{\circ}. \] 5. Recall the known values: \[ \cos 90^{\circ} = 0 \quad \text{and} \quad \sin 90^{\circ} = 1. \] 6. Substitute these values into the expression: \[ \cos \left(\theta-90^{\circ}\right) = \cos \theta \cdot 0 + \sin \theta \cdot 1 = \sin \theta. \] Thus, the simplified form is: \[ \cos \left(\theta-90^{\circ}\right) = \sin \theta. \]