Answer
$\dfrac{\csc x-\cot x}{\sec x-1}=\cot x$
Work Step by Step
$\dfrac{\csc x-\cot x}{\sec x-1}=\cot x$
On the left side of the equation, replace $\csc x$ by $\dfrac{1}{\sin x}$, $\cot x$ by $\dfrac{\cos x}{\sin x}$ and $\sec x$ by $\dfrac{1}{\cos x}$:
$\dfrac{\dfrac{1}{\sin x}-\dfrac{\cos x}{\sin x}}{\dfrac{1}{\cos x}-1}=\cot x$
Evaluate the subtractions indicated on the left side:
$\dfrac{\dfrac{\sin x-\sin x\cos x}{\sin^{2}x}}{\dfrac{1-\cos x}{\cos x}}=\cot x$
Evaluate the division on the left side:
$\dfrac{(\sin x-\sin x\cos x)(\cos x)}{(1-\cos x)(\sin^{2}x)}=\cot x$
Take out common factor $\sin x$ from $\sin x-\sin x\cos x$ and simplify:
$\dfrac{\sin x(1-\cos x)(\cos x)}{(1-\cos x)(\sin^{2}x)}=\cot x$
$\dfrac{\cos x}{\sin x}=\cot x$
Since $\dfrac{\cos x}{\sin x}=\cot x$, the identity is proved:
$\cot x=\cot x$
