# Simplifying $\frac{\cot(34^\circ)\sin(44^\circ)}{\sin(22^\circ)\sin(56^\circ)}$

$$\dfrac{\cot(34^\circ)\sin(44^\circ)}{\sin(22^\circ)\sin(56^\circ)}$$ Simplify given expression.

We know that $$\cot \alpha = \dfrac{\sin \alpha }{\cos \alpha }$$ and $$\sin(44^\circ) = 2\sin(22^\circ)\cos(22^\circ)$$

$$\dfrac{\cot(34^\circ) 2\sin(22^\circ)\cos(22^\circ)}{\sin(22^\circ)\sin(56^\circ)}$$

Cancelling similar terms out

$$\dfrac{\cot(34^\circ)2\cos (22^\circ)}{\sin(56^\circ)} = \dfrac{\dfrac{\cos(34^\circ)}{\sin(34^\circ)}2\cos (22^\circ)}{\sin(56^\circ)}$$

Using the identity $$\cos (90^\circ-\alpha ) = \sin (\alpha)$$

$$\dfrac{\dfrac{\sin(56^\circ)}{\sin(34^\circ)}2\cos (22^\circ)}{\sin(56^\circ)}=\dfrac {2\sin(56^\circ)\cos(22^\circ)}{\sin(34^\circ)\sin(56^\circ)} = \dfrac{2\cos (22^\circ)}{\sin(34^\circ)}$$

Since the degrees aren't the same, I could not proceed further.

$$\frac{2\sin 68^\circ}{\sin 34^\circ}=4\cos 34^\circ$$