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