Answer
See the full explanation below.
Work Step by Step
${{\left( \csc x-\cot x \right)}^{2}}$
Apply the reciprocal identity $\csc x=\frac{1}{\sin x}$, and quotient identity of trigonometry $\cot x=\frac{\cos x}{\sin x}$. Then the above expression can be further simplified as:
$\begin{align}
& {{\left( \csc x-\cot x \right)}^{2}}={{\left( \frac{1}{\sin x}-\frac{\cos x}{\sin x} \right)}^{2}} \\
& ={{\left( \frac{1-\cos x}{\sin x} \right)}^{2}} \\
& =\frac{{{\left( 1-\cos x \right)}^{2}}}{{{\sin }^{2}}x}
\end{align}$
Now, let us consider the right side of the given expression:
$\frac{1-\cos x}{1+\cos x}$
We multiply $\frac{1-\cos x}{1-\cos x}$ to the given expression:
$\begin{align}
& \frac{1-\cos x}{1+\cos x}=\frac{1-\cos x}{1+\cos x}.\frac{1-\cos x}{1-\cos x} \\
& =\frac{{{\left( 1-\cos x \right)}^{2}}}{1-{{\cos }^{2}}x}
\end{align}$
Apply the Pythagorean identity of trigonometry ${{\sin }^{2}}x=1-{{\cos }^{2}}x$, which comes out by solving ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Then the above expression can be further simplified as:
$\frac{{{\left( 1-\cos x \right)}^{2}}}{{{\sin }^{2}}x}$
The identity is verified because both sides are equal to $\frac{{{\left( 1-\cos x \right)}^{2}}}{{{\sin }^{2}}x}$.
Thus, the left side of the expression is equal to the right side, which is
${{\left( \csc x-\cot x \right)}^{2}}=\frac{1-\cos x}{1+\cos x}$.
