## Precalculus (6th Edition) Blitzer

Let us consider the left side of the given expression: $\frac{\cos \left( \alpha +\beta \right)}{\cos \left( \alpha -\beta \right)}$ By using the identities of trigonometry, $\cos \left( \alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta$ $\cos \left( \alpha -\beta \right)=\cos \alpha \cos \beta +sin\alpha \sin \beta$ , Now above provided expression can be further simplified as: $\frac{\cos \left( \alpha +\beta \right)}{\cos \left( \alpha -\beta \right)}=\frac{\cos \alpha \cos \beta -\sin \alpha \sin \beta }{\cos \alpha \cos \beta +sin\alpha \sin \beta }$ By dividing the numerator and denominator by the $\cos \,\alpha \cos \,\beta$, we get: \begin{align} & \frac{\cos \alpha \cos \beta -\sin \alpha \sin \beta }{\cos \alpha \cos \beta +sin\alpha \sin \beta }=\frac{\frac{\cos \alpha \cos \beta }{\cos \alpha \cos \beta }-\frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}{\frac{\cos \alpha \cos \beta }{\cos \alpha \cos \beta }+\frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }} \\ & =\frac{1-\frac{\sin \alpha }{\cos \alpha }\frac{\sin \beta }{\cos \beta }}{1+\frac{\sin \alpha }{\cos \alpha }\frac{\sin \beta }{\cos \beta }} \\ & =\frac{1-\tan \alpha \tan \beta }{1+\tan \alpha \tan \beta } \end{align} Hence, the left side of the given expression is equal to the right side, which is $\frac{\cos \left( \alpha +\beta \right)}{\cos \left( \alpha -\beta \right)}=\frac{1-\tan \alpha \tan \beta }{1+\tan \alpha \tan \beta }$.