Can $\frac{\sqrt{3}}{\sin20^{\circ}}-\frac{1}{\cos20^{\circ}}$ have two values? I would like to confirm a solution. The question goes as:
Show that $$\frac{\sqrt{3}}{\sin20^{\circ}}-\frac{1}{\cos20^{\circ}}=4$$
Firstly I combined the two terms to form something like: $$\dfrac{\sqrt{3}\cos20^{\circ} - \sin20^{\circ}}{\sin20^{\circ}\cos20^{\circ}}.$$
Clearly, the numerator is of the form $a\cos\theta+b\sin\theta$, and can thus be expressed as $R\sin(\theta+\alpha)$, where $R=\sqrt{a^2+b^2}$, and $\alpha=\arctan(\dfrac{a}{b})$. Following this method, I obtain something like this: $$\dfrac{2\sin(20^{\circ}-60^{\circ})}{\sin20^{\circ}\cos20^{\circ}},$$ which results in the expression to be equal to $-4$. But, we took $R$ to be the principal square root, which was $2$. If we take it to be the negative root, $-2$, it follows that the expression is equal to $4$. So, can this expression have two values? 
Edit: I would also like to mention how I computed the numerator:
Numerator: 
$\sqrt{3}\cos20^{\circ} - \sin20^{\circ}$.
Now, this can be represented as $R\sin(\theta+\alpha)$.
Computation of $R$: $R=\sqrt{a^2+b^2}\implies R=\sqrt{(3+1)}=2$
Computation of $\alpha$: $\tan\alpha=\dfrac{a}{b}\implies \tan\alpha=\dfrac{\sqrt3}{-1}\implies \alpha=-60^{\circ}$, because $\tan(-x)=-\tan x$, and in this case $\tan\alpha=-\sqrt3$, and $-\tan(60^{\circ})=-\sqrt3$, so $\alpha=-60^{\circ}$.
Hence, we conclude that numerator:$$2\sin(20+(-60))=2\sin(-40)=-2\sin(40^{\circ})$$

When $R=-2,-2\sin\theta=\sqrt3\iff\sin\theta=?,-2\cos\theta=1\iff\cos\theta=?$
Observe that $\theta$ lies in the third quadrant $\theta=(2n+1)180^\circ+60^\circ$ where $n$ is any integer
$$\sin\{20^\circ-(2n+1)180^\circ+60^\circ)\}=-\sin220^\circ$$
and $$\sin220^\circ=\sin(180+40)^\circ=-\sin40^\circ$$
