## Precalculus (6th Edition) Blitzer

We have the expression on the left side $\sec (\alpha +\beta )$, which can be simplified by using the reciprocal identity $\sec x=\frac{1}{\cos x}$. Thus, the expression can be written as: $\sec (\alpha +\beta )=\frac{1}{\cos \left( \alpha +\beta \right)}$ Then, use the sum identity: $\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y$ Then after applying the identity, the expression will be simplified as: \begin{align} & \sec (\alpha +\beta )=\frac{1}{\cos \left( \alpha +\beta \right)} \\ & =\frac{1}{\cos \alpha \cos \beta -\sin \alpha \sin \beta } \end{align} And multiplying and dividing the expression by $\frac{1}{\cos \alpha \cos \beta }$, we get: \begin{align} & \frac{1}{\cos \alpha \cos \beta -\sin \alpha \sin \beta }=\frac{1}{\cos \alpha \cos \beta -\sin \alpha \sin \beta }.\frac{\frac{1}{\cos \alpha \cos \beta }}{\frac{1}{\cos \alpha \cos \beta }} \\ & =\frac{\frac{1}{\cos \alpha \cos \beta }}{\frac{\cos \alpha \cos \beta -\sin \alpha \sin \beta }{\cos \alpha \cos \beta }} \\ & =\frac{\frac{1}{\cos \alpha \cos \beta }}{\frac{\cos \alpha \cos \beta }{\cos \alpha \cos \beta }-\frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }} \end{align} We know the reciprocal identity $\frac{1}{\cos x}=\sec x$ and quotient identity $\frac{\sin x}{\cos x}=\tan x$. Thus, applying the identities, we get $\frac{\frac{1}{\cos \alpha \cos \beta }}{\frac{\cos \alpha \cos \beta }{\cos \alpha \cos \beta }-\frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}=\frac{\sec \alpha \sec \beta }{1-\tan \alpha \tan \beta }$ Hence, the expression on the left-side is equal to the expression on the right-side.