Answer
The expression in terms of the function $\sec x\ \text{ and }\ \tan x$ is $-4\sec x\tan x$.
Work Step by Step
By multiplying the part of expression $\frac{1-\sin x}{1+\sin x}$ by $\frac{1-\sin x}{1-\sin x}$ and the other part of expression $\frac{1+\sin x}{1-\sin x}$ by $\frac{1+\sin x}{1+\sin x}$:
$\begin{align}
& \frac{1-\sin x}{1+\sin x}-\frac{1+\sin x}{1-\sin x}=\frac{1-\sin x}{1+\sin x}\times \frac{1-\sin x}{1-\sin x}-\frac{1+\sin x}{1-\sin x}\times \frac{1+\sin x}{1+\sin x} \\
& =\frac{1-\sin x-\sin x+{{\sin }^{2}}x}{1-{{\sin }^{2}}x}-\frac{1+\sin x+\sin x+{{\sin }^{2}}x}{1-{{\sin }^{2}}x} \\
& =\frac{1-2\sin x+{{\sin }^{2}}x-\left( 1+2\sin x+{{\sin }^{2}}x \right)}{1-{{\sin }^{2}}x} \\
& =\frac{1-2\sin x+{{\sin }^{2}}x-1-2\sin x-{{\sin }^{2}}x}{1-{{\sin }^{2}}x}
\end{align}$
Further solve:
$\frac{1-2\sin x+{{\sin }^{2}}x-1-2\sin x-{{\sin }^{2}}x}{1-{{\sin }^{2}}x}=\frac{-4\sin x}{1-{{\sin }^{2}}x}$
Now, apply the Pythagorean identity of trigonometry ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ , $1-si{{n}^{2}}x$ can be written as ${{\cos }^{2}}x$:
$\begin{align}
& \frac{-4\sin x}{1-{{\sin }^{2}}x}=\frac{-4\sin x}{{{\cos }^{2}}x} \\
& =-4\cdot \frac{1}{\cos x}\cdot \frac{\sin x}{\cos x}
\end{align}$
By using the reciprocal identity of trigonometry $\sec x=\frac{1}{\cos x}$ , and the quotient identity of trigonometry $\tan x=\frac{\sin x}{\cos x}$ , then the above expression can be further simplified as:
$\begin{align}
& -4\cdot \frac{1}{\cos x}\cdot \frac{\sin x}{\cos x}=-4\sec x\cdot \tan x \\
& =-4\sec x\tan x
\end{align}$
