<|system|>
Simplify $\frac{\csc \theta}{\sin \theta} - \frac{\cot \theta}{\tan \theta}.$</s>.
<|user|>
Can you answer based on context?
<|assistant|>
I know that $\csc \theta = \frac{1}{\sin \theta},$ so the first fraction becomes $\frac{1}{\sin \theta \cdot \sin \theta} = \frac{1}{\sin^2 \theta}.$ Similarly, I know that $\cot \theta = \frac{\cos \theta}{\sin \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta},$ so the second fraction becomes $\frac{\cos \theta}{\sin \theta} \cdot \frac{\cos \theta}{\sin \theta} = \frac{\cos^2 \theta}{\sin^2 \theta}.$ Now I can rewrite the expression as $\frac{1}{\sin^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1 - \cos^2 \theta}{\sin^2 \theta}.$ But I also know that $\sin^2 \theta + \cos^2 \theta = 1,$ so $1 - \cos^2 \theta = \sin^2 \theta.$ This means that the expression simplifies to $\frac{\sin^2 \theta}{\sin^2 \theta} = 1.$ 