Answer
$\dfrac{\cot\theta}{\csc\theta-\sin\theta}=\sec\theta$
Work Step by Step
$\dfrac{\cot\theta}{\csc\theta-\sin\theta}$
Substitute $\cot\theta$ with $\dfrac{\cos\theta}{\sin\theta}$ and $\csc\theta$ with $\dfrac{1}{\sin\theta}$:
$\dfrac{\cot\theta}{\csc\theta-\sin\theta}=\dfrac{\dfrac{\cos\theta}{\sin\theta}}{\dfrac{1}{\sin\theta}-\sin\theta}=...$
Evaluate the difference present in the denominator and simplify:
$...=\dfrac{\dfrac{\cos\theta}{\sin\theta}}{\dfrac{1-\sin^{2}\theta}{\sin\theta}}=\dfrac{\cos\theta}{1-\sin^{2}\theta}=...$
We know that $\sin^{2}\theta+\cos^{2}\theta=1$, so if we solve for $\cos^{2}\theta$ here, we will get the expression present in the denominator. So, substituting $1-\sin^{2}\theta$ with $\cos^{2}\theta$, the expression becomes:
$...=\dfrac{\cos\theta}{\cos^{2}\theta}=\dfrac{1}{\cos\theta}=\sec\theta$
