Trigonometry problem $\sin100^\circ+\cos70^\circ\over\cos80^\circ-\cos20^\circ$ 
What is the value of:
  $$\sin100^\circ+\cos70^\circ\over\cos80^\circ-\cos20^\circ$$

I've done trigonometry in my earlier years of high school but I forgot a lot of rules. This is where I'm stuck on this problem:
$\large{{\sin100^\circ+\cos70^\circ\over\cos80^\circ-\cos20^\circ}=\\{\sin(90^\circ+10^\circ)+\cos(60^\circ+10^\circ)\over\cos(90^\circ-10^\circ)-\cos(30^\circ+10^\circ)}=\\{\sin90^\circ\cos10^\circ+\cos90^\circ\sin10^\circ+\cos60^\circ\cos10^\circ-\sin60^\circ\sin10^\circ\over\cos90^\circ\cos10^\circ+\sin90^\circ\sin10^\circ-\cos30^\circ\cos10^\circ+\sin30^\circ\sin10^\circ}=\\{\cos10^\circ+{1\over2}\cos10^\circ-{\sqrt3\over2}\sin10^\circ\over\sin10^\circ-{\sqrt3\over2}\cos10^\circ+{1\over2}\sin10^\circ}=\\{{3\over2}\cos10^\circ-{\sqrt3\over2}\sin10^\circ\over{3\over2}\sin10^\circ-{\sqrt3\over2}\cos10^\circ}}$
Not sure what I should do further with this.
 A: It is:
$$\frac{\sin 100^\circ +\cos 70^\circ}{\cos 80^\circ-\cos 20^\circ}=\frac{\sin 80^\circ +\sin 20^\circ}{\cos 80^\circ-\cos 20^\circ}=\frac{2\sin 50^\circ\cos 30^\circ}{-2\sin 50^\circ\sin 30^\circ}=-\sqrt{3}.$$
A: Given $$\dfrac{\sin100+\cos100}{\cos80-\cos20}$$
Now to solve the denominator use the formula $\cos A-\cos B=-2\sin\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)$
$\cos80-\cos20=-2\sin\left(\dfrac{80+20}{20}\right)\sin\left(\dfrac{80-20}{2}\right)$
$$=\dfrac{\sin100+\cos70}{-2\sin\left(\dfrac{80+20}{20}\right)\sin\left(\dfrac{80-20}{2}\right)}$$
$$=-\dfrac{\sin100+\cos70}{\sin50}$$
Now to solve the numerator use the identity $\sin A+\sin B=2\cos\left(\dfrac{A-B}{2}\right)\sin\left(\dfrac{A+B}{2}\right)$
$\sin100+\sin20=2\cos\left(\dfrac{100-20}{2}\right)\sin\left(\dfrac{100+20}{2}\right)$
$$=-\dfrac{2\cos\left(\dfrac{100-20}{2}\right)\sin\left(\dfrac{100+20}{2}\right)}{\sin50}$$
$$=-\dfrac{2\cos40\sin60}{\sin50}$$
$$=-\dfrac{2\cos40\cdot\dfrac{\sqrt{3}}{2}}{\sin50}$$
$$-\dfrac{\sqrt{3}\sin(90-40)}{\sin50}=-\sqrt{3}$$
Therefore, $$\dfrac{\sin100+\cos100}{\cos80-\cos20}=-\sqrt{3}$$
A: Since $\cos x=\sin(90-x)$, $$\frac{\sin100+\cos70}{\cos80-\cos20}=\frac{2\sin\frac{100+20}2\cos\frac{100-20}2}{2\sin\frac{80+20}2\sin\frac{20-80}2}=-\frac{\sin60\cos40}{\sin50\sin30}=-\frac{\frac{\sqrt3}2\sin50}{\frac12\sin50}=-\sqrt3$$ using $$\sin  x+\sin y=2\sin\frac{x+y}2\cos\frac{x-y}2\\\cos x-\cos y=-2\sin\frac{x+y}2\sin\frac{x-y}2$$
