If $\tan {20}^{\circ }=k$ then $\frac{\tan {250}^{\circ }+\tan {340}^{\circ }}{\tan {20}^{\circ }-\tan {110}^{\circ }}=$

$\tan {20}^{\circ }=k$

$\frac{\tan {250}^{\circ }+\tan {340}^{\circ }}{\tan {20}^{\circ }-\tan {110}^{\circ }}$

$=\frac{\tan ({270}^{\circ }-{20}^{\circ })+\tan ({360}^{\circ }-{20}^{\circ })}{\tan {20}^{\circ }-\tan ({90}^{\circ }+{20}^{\circ })}$

$=\frac{\cot {20}^{\circ }-\tan {20}^{\circ }}{\tan {20}^{\circ }+\cot {20}^{\circ }}$

$=\frac{{\displaystyle \frac{1}{k}}-k}{k+{\displaystyle \frac{1}{k}}}$

$=\frac{1-k^2}{1+k^2}$