Answer
$$\frac{\csc\theta+\cot\theta}{\tan\theta+\sin\theta}=\cot\theta\csc\theta$$
We simplify the left side first, and the trigonometric expression is an identity.
Work Step by Step
$$\frac{\csc\theta+\cot\theta}{\tan\theta+\sin\theta}=\cot\theta\csc\theta$$
The left side is more complicated, which means we would deal with it first.
$$A=\frac{\csc\theta+\cot\theta}{\tan\theta+\sin\theta}$$
We would use the following identities:
$$\csc\theta=\frac{1}{\sin\theta}\hspace{1cm}\cot\theta=\frac{\cos\theta}{\sin\theta}\hspace{1cm}\tan\theta=\frac{\sin\theta}{\cos\theta}$$
Apply them into $A$:
$$A=\frac{\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}}{\frac{\sin\theta}{\cos\theta}+\sin\theta}$$
$$A=\frac{\frac{1+\cos\theta}{\sin\theta}}{\frac{\sin\theta+\sin\theta\cos\theta}{\cos\theta}}$$
$$A=\frac{1+\cos\theta}{\sin\theta}\times\frac{\cos\theta}{\sin\theta+\sin\theta\cos\theta}$$
$$A=\frac{1+\cos\theta}{\sin\theta}\times\frac{\cos\theta}{\sin\theta(1+\cos\theta)}$$
$$A=\frac{\cos\theta}{\sin^2\theta}$$
$$A=\frac{\cos\theta}{\sin\theta}\times\frac{1}{\sin\theta}$$
Also, from the following identities:
$$\frac{\cos\theta}{\sin\theta}=\cot\theta\hspace{1.5cm}\frac{1}{\sin\theta}=\csc\theta$$
Therefore, $$A=\cot\theta\csc\theta$$
The trigonometric expression is therefore proved.
