Answer
$\dfrac{\cos\theta}{1-\sin\theta}=\dfrac{\sin\theta-\csc\theta}{\cos\theta-\cot\theta}$
Work Step by Step
$\dfrac{\cos\theta}{1-\sin\theta}=\dfrac{\sin\theta-\csc\theta}{\cos\theta-\cot\theta}$
On the right side of the equation, replace $\csc\theta$ by $\dfrac{1}{\sin\theta}$ and $\cot\theta$ by $\dfrac{\cos\theta}{\sin\theta}$:
$\dfrac{\cos\theta}{1-\sin\theta}=\dfrac{\sin\theta-\dfrac{1}{\sin\theta}}{\cos\theta-\dfrac{\cos\theta}{\sin\theta}}$
Evaluate the subtractions on the right side:
$\dfrac{\cos\theta}{1-\sin\theta}=\dfrac{\dfrac{\sin^{2}\theta-1}{\sin\theta}}{\dfrac{\sin\theta\cos\theta-\cos\theta}{\sin\theta}}$
Evaluate the division on the right:
$\dfrac{\cos\theta}{1-\sin\theta}=\dfrac{\sin\theta(\sin^{2}\theta-1)}{\sin\theta(\sin\theta\cos\theta-\cos\theta)}$
$\dfrac{\cos\theta}{1-\sin\theta}=\dfrac{\sin^{2}\theta-1}{\sin\theta\cos\theta-\cos\theta}$
Take out common factor $\cos\theta$ from the denominator on the right:
$\dfrac{\cos\theta}{1-\sin\theta}=\dfrac{\sin^{2}\theta-1}{\cos\theta(\sin\theta-1)}$
Change the sign of the numerator and the sign of the denominator on the right:
$\dfrac{\cos\theta}{1-\sin\theta}=\dfrac{1-\sin^{2}\theta}{\cos\theta(1-\sin\theta)}$
Replace $1-\sin^{2}\theta$ by $\cos^{2}\theta$:
$\dfrac{\cos\theta}{1-\sin\theta}=\dfrac{\cos^{2}\theta}{\cos\theta(1-\sin\theta)}$
Simplify and the identity will be proved:
$\dfrac{\cos\theta}{1-\sin\theta}=\dfrac{\cos\theta}{1-\sin\theta}$
