How to simplify this trig expression?

I have given this: $$\frac{1}{8} \cot \left(\frac{t}{2}\right) \csc ^2\left(\frac{t}{2}\right) \csc (t)$$

And I have to show that it is that: $$\frac{1}{16} \csc ^4\left(\frac{t}{2}\right)$$

I tried to use the doule angle formula on the $\sin(t)$, but I ended up with $\frac 14 \cot^2\left(\frac t2\right)$. Mathematica gets it right, but it does not tell me how it simplified it.

How do I do this?

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$$\frac{1}{8} \cot \left(\frac{t}{2}\right) \csc ^2\left(\frac{t}{2}\right) \csc (t)$$ $$=\frac{1}{8} \cot \left(\frac{t}{2}\right) \csc ^2\left(\frac{t}{2}\right) \frac{1}{\sin t}$$ $$=\frac{1}{8} \frac{\cos \left(\frac{t}{2}\right)}{\sin \left(\frac{t}{2}\right)} \csc ^2\left(\frac{t}{2}\right)\frac{1}{2\sin(\frac{t}{2})\cos(\frac{t}{2})}$$ $$=\frac{1}{16}\csc^4 (\frac{t}{2})$$ ${}{}{}$