Answer
See the full explanation below.
Work Step by Step
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.
