## Precalculus (6th Edition) Blitzer

Left side, $\frac{\sin \left( x-y \right)}{\cos x\cos y}+\frac{\sin \left( y-z \right)}{\cos y\cos z}+\frac{\sin \left( z-x \right)}{\cos z\cos x}$ By using trigonometric identity, $\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta$ Left side can be written as \begin{align} & \frac{\sin \left( x-y \right)}{\cos x\cos y}+\frac{\sin \left( y-z \right)}{\cos y\cos z}+\frac{\sin \left( z-x \right)}{\cos z\cos x}=\frac{\sin x\cos y-\cos x\sin y}{\cos x\cos y}+ \\ & \frac{\sin y\cos z-\cos y\sin z}{\cos y\cos z}+\frac{\sin z\cos x-\cos z\sin x}{\cos z\cos x} \\ & =\frac{\sin x\cos y}{\cos x\cos y}-\frac{\cos x\sin y}{\cos x\cos y}+\frac{\sin y\cos z}{\cos y\cos z}-\frac{\cos y\sin z}{\cos y\cos z}+ \\ & \frac{\sin z\cos x}{\cos z\cos x}-\frac{\cos z\sin x}{\cos z\cos x} \end{align} Then, arranging above equation \begin{align} & \frac{\sin x\cos y}{\cos x\cos y}-\frac{\cos x\sin y}{\cos x\cos y}+\frac{\sin y\cos z}{\cos y\cos z}-\frac{\cos y\sin z}{\cos y\cos z}+\frac{\sin z\cos x}{\cos z\cos x}-\frac{\cos z\sin x}{\cos z\cos x}=\frac{\sin x}{\cos x}-\frac{\sin y}{\cos y}+ \\ & \frac{\sin y}{\cos y}-\frac{\sin z}{\cos z} \\ & +\frac{\sin z}{\cos z}-\frac{\sin x}{\cos x} \\ & =\frac{\sin x}{\cos x}-\frac{\sin x}{\cos x} \\ & +\frac{\sin y}{\cos y}-\frac{\sin y}{\cos y} \\ & +\frac{\sin z}{\cos z}-\frac{\sin z}{\cos z} \\ & =0 \end{align} Thus, it is verified that the left side and right side of the given expression are equal.