Answer
$\dfrac{\cos^{2}x}{1-\sin x}=\dfrac{\cos x}{\sec x-\tan x}$
Work Step by Step
$\dfrac{\cos^{2}x}{1-\sin x}=\dfrac{\cos x}{\sec x-\tan x}$
On the right side, substitute $\sec x$ by $\dfrac{1}{\cos x}$ and $\tan x$ by $\dfrac{\sin x}{\cos x}$:
$\dfrac{\cos^{2}x}{1-\sin x}=\dfrac{\cos x}{\dfrac{1}{\cos x}-\dfrac{\sin x}{\cos x}}$
Evaluate the subtraction on the denominator in the right side:
$\dfrac{\cos^{2}x}{1-\sin x}=\dfrac{\cos x}{\dfrac{1-\sin x}{\cos x}}$
Evaluate the division on the right side and the identity will be proved:
$\dfrac{\cos^{2}x}{1-\sin x}=\dfrac{\cos^{2}x}{1-\sin x}$
