Answer
See the proof below.
Work Step by Step
Consider the left side of the equation $\frac{1+\sin x}{1-\sin x}-\frac{1-\sin x}{1+\sin x}$.
At first, take the least common multiple of the denominator, $\begin{align}
& =\frac{\left( 1+\sin x \right)\left( 1+\sin x \right)-\left( 1-\sin x \right)\left( 1-\sin x \right)}{\left( 1-\sin x \right)\times \left( 1+\sin x \right)} \\
& =\frac{\left( 1+{{\sin }^{2}}x+2\sin x \right)-\left( 1+{{\sin }^{2}}x-2\sin x \right)}{\left( 1-{{\sin }^{2}}x \right)} \\
\end{align}$
Then reduce the above equation in simplified form
$\begin{align}
& =4\frac{\sin x}{{{\cos }^{2}}x} \\
& =4\frac{\sin x}{\cos x\times \cos x} \\
\end{align}$
Also, $\frac{\sin x}{\cos x}=\tan x\,\text{ and }\,\frac{1}{\cos x}=\sec x $
The simplified form of equation is:
$\begin{align}
& =4\frac{\sin x}{\cos x}\times \frac{1}{\cos x} \\
& =4\tan x\sec x \\
\end{align}$
Hence, the identity is verified.
