Answer
$$\frac{1+\sin\theta}{\cot^2\theta}=\frac{\sin\theta}{\csc\theta-1}$$
The equation is an identity.
Work Step by Step
$$\frac{1+\sin\theta}{\cot^2\theta}=\frac{\sin\theta}{\csc\theta-1}$$
We examine from the left side first.
$$A=\frac{1+\sin\theta}{\cot^2\theta}$$
$\cot\theta$ can be written according to Quotient Identity, in which $\cot\theta=\frac{\cos\theta}{\sin\theta}$
$$A=\frac{1+\sin\theta}{\frac{\cos^2\theta}{\sin^2\theta}}$$
$$A=\frac{\sin^2\theta(1+\sin\theta)}{\cos^2\theta}$$
Next, we write $\cos^2\theta=1-\sin^2\theta=(1-\sin\theta)(1+\sin\theta)$, following the Pythagorean Identity.
$$A=\frac{\sin^2\theta(1+\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}$$
$$A=\frac{\sin^2\theta}{1-\sin\theta}$$
The left side can be temporarily stopped being examined to switch to the right side.
$$B=\frac{\sin\theta}{\csc\theta-1}$$
$\csc\theta$ can be written according to Reciprocal Identity: $\csc\theta=\frac{1}{\sin\theta}$
$$B=\frac{\sin\theta}{\frac{1}{\sin\theta}-1}$$
$$B=\frac{\sin\theta}{\frac{1-\sin\theta}{\sin\theta}}$$
$$B=\frac{\sin^2\theta}{1-\sin\theta}$$
Thus, $A=B$. 2 sides are equal, so the equation is verified to be an identity.
