What is the exact value of $\frac{1}{\tan20^{\circ}}-\frac{1}{\sin80^{\circ}}$? 
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.
 A: Colored for easier reading.  \begin{eqnarray}\frac{1}{\tan20^{\circ}}-\frac{1}{\sin80^{\circ}}&=&\frac{\cos20^{\circ}}{\sin20^{\circ}}-\frac{1}{\cos10^{\circ}}\\
&=&\color{red}{\frac{\cos20^{\circ}}{2\sin10^{\circ}\cos10^{\circ}}-\frac{1}{\cos10^{\circ}}}\\
&=&\frac{\cos20^{\circ}-2\sin10^{\circ}}{2\sin10^{\circ}\cos10^{\circ}}\\
&=&\color{red}{2\frac{\cos60^{\circ}\cos20^{\circ}-\sin10^{\circ}}{\sin20^{\circ}}}\\
&=&\frac{\cos80^{\circ}+\cos40^{\circ}-2\sin10^{\circ}}{\sin20^{\circ}}\\
&=&\color{red}{\frac{\cos40^{\circ}-\cos80^{\circ}}{\sin20^{\circ}}}\\
&=&\frac{\cos40^{\circ}+\cos100^{\circ}}{\sin20^{\circ}}\\
&=&\color{red}{\frac{2\cos70^{\circ}\cos30^{\circ}}{\sin20^{\circ}}}\\
&=&2\cos30^{\circ}=\sqrt{3}
\end{eqnarray}
A: 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.
