Answer
See the explanation below.
Work Step by Step
We use the trigonometric identities: $2\sin x\cos x=\sin 2x$ and $\cos 2x=2{{\cos }^{2}}x-1$.
Now, the left side can be written as:
$\begin{align}
& \frac{\sin 2x}{\sin x}-\frac{\cos 2x}{\cos x}=\frac{2\sin x\cos x}{\sin x}-\frac{\left( 2{{\cos }^{2}}x-1 \right)}{\cos x} \\
& =2\cos x-\frac{2{{\cos }^{2}}x}{\cos x}+\frac{1}{\cos x} \\
& =2\cos x-2\cos x+\sec x \\
& =\sec x
\end{align}$
Thus, the left side is equal to $\sec x$.
