Simplfying a trigonometric expression

Question

I would like to simplify:

$$\frac{\cos^2(80)+5\sin^2(80)-3}{\cos(50)}$$

By using the fact that $\sin^2(\theta) + \cos^2(\theta) = 1$,

$$\frac{\cos^2(80)+5\sin^2(80)-3}{\cos(50)} = \frac{\cos^2(80)+5(1-\cos^2(80))-3}{\cos(50)} = -2\biggr(\frac{2\cos^2(80)-1}{\cos(50)}\biggr)$$

Since $2\cos^2(80)-1 = \cos(160)$,

$$-2\biggr(\frac{2\cos^2(80)-1}{\cos(50)}\biggr) = -2\frac{\cos(160)}{\cos(50)}$$

But I am not sure how we could simplify this further.

Answer 1

Since $\cos(\pi-\theta)=-\cos\theta$, if the angles are given in degrees $\cos(180^{\circ}-\theta)=-\cos \theta$, it follows

\begin{align*} -2\dfrac{\cos 160^{\circ}}{\cos 50^{\circ}}&=\dfrac{2\cos 20^{\circ}}{\cos 50^{\circ}}\\ &=\dfrac{2\cos 20^{\circ}}{\sin(90^{\circ}-50^{\circ})}\\ &=\dfrac{2\cos 20^{\circ}}{\sin 40^{\circ}}\\ &=\dfrac{2\cos 20^{\circ}}{2\sin 20^{\circ}\cos 20^{\circ}}\\ &=\dfrac1{\sin 20^{\circ}} \end{align*}