If $\tan {20}^0 =\mathrm{\lambda },$ then $\frac{\tan {160}^0 -\tan {110}^0}{1+\left(\tan {160}^0\right)\left(\tan {110}^0\right)}=$

$\begin{array}{rcl}\tan {20}^0& =& \mathrm{\lambda }\\ & & \frac{\tan {160}^0-\tan {110}^0}{1+\tan {160}^0 \tan {110}^0}\\ & =& \frac{\tan \left({180}^0-{20}^0\right)-\tan \left({90}^0+{20}^0\right)}{1+\tan \left({180}^0-{20}^0\right)\tan \left({90}^0+{20}^0\right)}\\ & =& \frac{-\tan {20}^0+\cot {20}^0}{1-\tan {20}^0\left(-\cot {20}^0\right)}\\ & =& \frac{-\mathrm{\lambda }+{\displaystyle \frac{1}{\mathrm{\lambda }}}}{1+1}\\ & =& \frac{1-{\mathrm{\lambda }}^2}{2\mathrm{\lambda }}\end{array}$