Given that \( \cos (A-B)=\cos A \cdot \cos B+\sin A \cdot \sin B \) use \( \cos (A-B) \) to prove that \( \cos (A+B)=\cos A-\cos B-\sin A \cdot \sin B \) Simplify : \( \frac{\cos (\theta-90 \%}{\sin (90+\theta) \tan \left(720^{\circ}-\theta\right)} \)

To start, we can express \( \cos(A+B) \) using the angle subtraction formula. Recall that \( \cos(A+B) = \cos(A - (-B)) \), which allows us to leverage the known formula \( \cos(A-B) \). Thus, we substitute: \[ \cos (A+B) = \cos A \cdot \cos(-B) + \sin A \cdot \sin(-B) = \cos A \cdot \cos B - \sin A \cdot \sin B \] Notice that \( \cos(-B) = \cos B \) and \( \sin(-B) = -\sin B \). Hence, the correct expression for \( \cos(A+B) \) is: \[ \cos(A+B) = \cos A \cdot \cos B - \sin A \cdot \sin B \] Next, we simplify the given expression. We first recognize \( \cos(\theta - 90^\circ) = \sin(\theta) \) and \( \sin(90 + \theta) = \cos(\theta) \). Thus, we rewrite the expression as follows: \[ \frac{\sin \theta}{\sin(90+\theta) \tan(720^\circ - \theta)} = \frac{\sin \theta}{\cos \theta \cdot \tan(720^\circ - \theta)} \] Now, using the identity \( \tan(720^\circ - \theta) = -\tan(\theta) \), we proceed: \[ = \frac{\sin \theta}{\cos \theta \cdot (-\tan \theta)} = \frac{\sin \theta}{\cos \theta \cdot (-\frac{\sin \theta}{\cos \theta})} = -\frac{\sin^2 \theta}{\cos^2 \theta} = -\tan^2 \theta \] So, the simplified form is: \[ -\tan^2 \theta \]