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


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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.