What is the value of 40^ - 140^ 80^ + 20^ ?

Answer: Option B

$$\begin{array}{rl}& \frac{\cos {40}^{\circ }-\cos {140}^{\circ }}{\sin {80}^{\circ }+\sin {20}^{\circ }}\\ & =\frac{-2\sin \left(\frac{{40}^{\circ }+{140}^{\circ }}{2}\right)\sin \left(\frac{{40}^{\circ }-{140}^{\circ }}{2}\right)}{2\sin \left(\frac{{80}^{\circ }+{20}^{\circ }}{2}\right)\cos \left(\frac{{80}^{\circ }-{20}^{\circ }}{2}\right)}\\ & =\frac{-2\sin \left(\frac{{180}^{\circ }}{2}\right)\sin \left(-\frac{{100}^{\circ }}{2}\right)}{2\sin \left(\frac{{100}^{\circ }}{2}\right)\cos \left(\frac{{60}^{\circ }}{2}\right)}\\ & =\frac{-2\sin {90}^{\circ }\times \sin \left(-{50}^{\circ }\right)}{2\sin {50}^{\circ }\times \cos {30}^{\circ }}\\ & =\frac{-2\sin {90}^{\circ }\times \left(-\sin {50}^{\circ }\right)}{2\sin {50}^{\circ }\times \cos {30}^{\circ }}\\ & =\frac{2\sin {90}^{\circ }\times \sin {50}^{\circ }}{2\sin {50}^{\circ }\times \cos {30}^{\circ }}\\ & =\frac{\sin {90}^{\circ }}{\cos {30}^{\circ }}\\ & =\frac{1}{\frac{\sqrt{3}}{2}}\\ & =\frac{2}{\sqrt{3}}\end{array}$$