What is the exact value of $\frac{1}{\tan20^{\circ}}-\frac{1}{\sin80^{\circ}}$?

Question

What is the exact value of $$\frac{1}{\tan20^{\circ}}-\frac{1}{\color{red}\sin80^{\circ}}?$$

By putting the above into a calculator, I get $\sqrt{3}$, but I cannot seem to be able to do it algebraically.

Answer 1

In any case for computing the result you can use the following trigonometric identities:

$\tan(80^{\circ})=\tan(\frac{\pi}{3}+20^{\circ})=\frac{\tan(\frac{\pi}{3})+\tan(20^{\circ})}{1-\tan(\frac{\pi}{3})\tan(20^{\circ})}=\frac{\sqrt{3}+\tan(20^{\circ})}{1-\sqrt{3}\tan(20^{\circ})}$

Now, let's target $\tan(20^{\circ})$ to find also the value for $\tan(80^{\circ})$. We can write:

$\tan(60^{\circ})=\tan(20^{\circ}+20^{\circ}+20^{\circ})=\frac{3\tan(20^{\circ})-\tan^3(20^{\circ})}{1-3\tan^2(20^{\circ})}$

Setting $x=\tan(20^{\circ})$ in the last equation we obtain: $\sqrt{3}=\frac{3x-x^3}{1-3x^2}$. And hence (assuming $1-3x\ne 0$):

$x^3-3\sqrt{3}x^2-3x+\sqrt{3}=0$

I think you can go from here.