Prove that $\ \ \ \ \ \ \ \dfrac{sin 20^\circ}{cos 20^\circ-2\cdot sin 10^\circ}=tan 30^\circ\ \ \ \ \ \ \ \ \ $ Solution
Proof: Because $$ sin 10^\circ-sin 10^\circ=0,\ \ \ sin^2 20^\circ+cos^2 20^\circ=1,\ \ \ sin 30^\circ =\frac{1}{2}$$
We have $$(sin^2 20^\circ+cos^2 20^\circ)\cdot sin 10^\circ-2\cdot sin 10^\circ\cdot sin 30^\circ=0$$
Adding $\ sin 20^\circ\cdot cos 20^\circ\cdot cos 10^\circ\ $ to both sides of the above equation, we have:
$$ sin 20^\circ\cdot cos 20^\circ\cdot cos 10^\circ + (sin^2 20^\circ+cos^2 20^\circ)\cdot sin 10^\circ-2\cdot sin 10^\circ\cdot sin 30^\circ$$
$$=sin 20^\circ\cdot cos 20^\circ\cdot cos 10^\circ $$
Moving the term $\ sin^2 20^\circ\cdot sin 10^\circ\ $ from the left side to the right side of the above equation, we have:
$$ sin 20^\circ\cdot cos 20^\circ\cdot cos 10^\circ + cos^2 20^\circ\cdot sin 10^\circ-2\cdot sin 10^\circ\cdot sin 30^\circ$$
$$=sin 20^\circ\cdot cos 20^\circ\cdot cos 10^\circ-sin^2 20^\circ\cdot sin 10^\circ$$
Combining the first two terms on the left side and two terms on the right side of the above equation, we have:
$$ cos 20^\circ\cdot(sin 20^\circ\cdot cos 10^\circ+cos 20^\circ\cdot sin 10^\circ)-2\cdot sin 10^\circ\cdot sin 30^\circ$$
$$=sin 20^\circ\cdot(cos 20^\circ\cdot cos 10^\circ-sin 20^\circ\cdot sin 10^\circ)$$
Because $$ sin 30^\circ=sin(20^\circ+10^\circ)=sin 20^\circ\cdot cos 10^\circ+cos 20^\circ\cdot sin 10^\circ $$
$$ cos 30^\circ=cos(20^\circ+10^\circ)=cos 20^\circ\cdot cos 10^\circ-sin 20^\circ\cdot sin 10^\circ$$
We have: $$ cos 20^\circ\cdot sin 30^\circ-2\cdot sin 10^\circ\cdot sin 30^\circ=sin 20^\circ\cdot cos 30^\circ$$
Finally, $$\boxed{\dfrac{sin 20^\circ}{cos 20^\circ-2\cdot sin 10^\circ}=\dfrac{sin 30^\circ}{cos 30^\circ}=tan 30^\circ}$$
