Prove $\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$
So, LS= $\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}$ $\left(\dfrac{1}{\sin\theta}\cdot \dfrac{\tan\theta}{1}\right)-\left(\dfrac{1}{\tan\theta}\cdot \dfrac{\sin\theta}{1}\right)$ $\dfrac{\tan\theta}{\sin\theta}-\dfrac{\sin\theta}{\tan\theta}$ Now, considering the fact that I must have a common denominator to subtract, would this be correct:
$\dfrac{\tan^2\theta}{\tan\theta\sin\theta}-\dfrac{\sin^2\theta}{\tan\theta\sin\theta}\Rightarrow \dfrac{\tan^2\theta-\sin^2\theta}{\tan\theta\sin\theta}$ I feel like I'm close to the answer because the denominator is the RS of the OP. Please help. Do not give me the answer.