Answer
See the explanation below.
Work Step by Step
$\frac{\sin x+\cos x}{\sin x}-\frac{\cos x-\sin x}{\cos x}$
By expressing the terms in the numerator and denominator with the least common denominator $\sin x\cos x$, we get:
$\begin{align}
& \frac{\sin x+\cos x}{\sin x}-\frac{\cos x-\sin x}{\cos x}=\frac{\cos x\left( \sin x+\cos x \right)-\sin x\left( \cos x-\sin x \right)}{\sin x\cos x} \\
& =\frac{\cos x\sin x+{{\cos }^{2}}x-\sin x\cos x+{{\sin }^{2}}x}{\sin x\cos x} \\
& =\frac{{{\cos }^{2}}x+{{\sin }^{2}}x}{\sin x\cos x}
\end{align}$
Apply the Pythagorean identity of trigonometry ${{\sin }^{2}}x+{{\cos }^{2}}x=1$; then, the above expression can be further simplified as:
$\frac{{{\cos }^{2}}x+{{\sin }^{2}}x}{\sin x\cos x}=\frac{1}{\sin x\cos x}$
Then, rewrite the above expression as:
$\frac{1}{\sin x\cos x}=\frac{1}{\sin x}.\frac{1}{\cos x}$
By using the reciprocal identity of trigonometry $\csc x=\frac{1}{\sin x}$ and $\sec x=\frac{1}{\cos x}$ , and now the above expression can be further simplified as:
$\begin{align}
& \frac{1}{\sin x}.\frac{1}{\cos x}=\csc x.\sec x \\
& =\csc x\sec x \\
& =\sec x\csc x
\end{align}$
Thus, the left side of the expression is equal to the right side, which is
$\frac{\sin x+\cos x}{\sin x}-\frac{\cos x-\sin x}{\cos x}=\sec x\csc x$.
