If tan40° = α, then find 320^ - 310^ 1 + 320^ 310^

Answer: Option C

$$\begin{array}{rl}& \frac{\tan {320}^{\circ }-\tan {310}^{\circ }}{1+\tan {320}^{\circ }\tan {310}^{\circ }}\\ & =\frac{\tan \left({360}^{\circ }-{40}^{\circ }\right)-\tan \left({270}^{\circ }+{40}^{\circ }\right)}{1+\tan \left({360}^{\circ }-{40}^{\circ }\right)\tan \left({270}^{\circ }+{40}^{\circ }\right)}\\ & =\frac{-\tan {40}^{\circ }+\cot {40}^{\circ }}{1+\tan {40}^{\circ }\times \cot {40}^{\circ }}\\ & =\frac{\cot {40}^{\circ }-\tan {40}^{\circ }}{1+1}\\ & =\frac{\frac{1}{\alpha }-\alpha }{2}\\ & =\frac{1-{\alpha }^2}{2\alpha }\end{array}$$