Answer
$\dfrac{1+\sin x}{1-\sin x}-\dfrac{1-\sin x}{1+\sin x}=4\tan x\sec x$
Work Step by Step
$\dfrac{1+\sin x}{1-\sin x}-\dfrac{1-\sin x}{1+\sin x}=4\tan x\sec x$
Evaluate the subtraction on the left side and simplify:
$\dfrac{(1+\sin x)^{2}-(1-\sin x)^{2}}{(1-\sin x)(1+\sin x)}=4\tan x\sec x$
$\dfrac{(1+2\sin x+\sin^{2}x)-(1-2\sin x+\sin^{2}x)}{1-\sin^{2}x}=4\tan x\sec x$
$\dfrac{1+2\sin x+\sin^{2}x-1+2\sin x-\sin^{2}x}{1-\sin^{2}x}=4\tan x\sec x$
$\dfrac{4\sin x}{1-\sin^{2}x}=4\tan x\sec x$
Replace $1-\sin^{2}x$ by $\cos^{2}x$:
$\dfrac{4\sin x}{\cos^{2}x}=4\tan x\sec x$
Rewrite the left side as $4\cdot\dfrac{\sin x}{\cos x}\cdot\dfrac{1}{\cos x}$:
$4\cdot\dfrac{\sin x}{\cos x}\cdot\dfrac{1}{\cos x}=4\tan x\sec x$
Since $\dfrac{\sin x}{\cos x}=\tan x$ and $\dfrac{1}{\cos x}=\sec x$, the identity is proved.
$4\tan x\sec x=4\tan x\sec x$
