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

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

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  • $\begingroup$ It was unbelieveable mistake. Thanks for reminding! $\endgroup$ – Hamilton Dec 1 '18 at 17:02

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