Prove the following:
46. $\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$
I got as far as
Right Side: $\tan\theta\sin\theta$ to $\dfrac{\sin\theta}{\cos\theta}\dfrac{\sin\theta}{1}$ and then; $\dfrac{\sin^2\theta}{\cos\theta}$
Left Side: $\begin{align*} \dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta} &= \dfrac{\frac{1}{\sin^2\theta}-{\frac{\cos^2\theta}{\sin^2\theta}}}{\frac{\cos\theta}{\sin\theta}-{\frac{1}{\sin^2\theta}}}\\ &= \dfrac{\frac{1-\cos^2\theta}{\sin^2\theta}}{\frac{\cos\theta}{\sin^2\theta}}\\ &= \dfrac{\frac{\sin^2\theta}{\sin^2\theta}}{\frac{\cos\theta}{\sin^2\theta}}\\ &= \frac{1}{\frac{\cos\theta}{\sin^2\theta}}\\ &= \frac{\sin^2\theta}{\cos\theta} \end{align*}$ Thanks a lot!