Answer:
Option D
Explanation:
Given, $\tan 20^{0}= \lambda$
$\therefore$ $\frac{\tan 160^{0}-\tan 110^{0}}{1+ (\tan 160^{0})(\tan 110^{0}) }$
=$\frac{\tan (180^{0}-20^{0})-\tan (90^{0}+20^{0})}{1+ (\tan( 180^{0}-20^{0})(\tan 90^{0}+20^{0}) }$
$=\frac{-\tan 20^{0}+\cot 20^{0}}{1+ \tan 20^{0} \cot 20^{0}}$
$[ \because \tan (180^{0}-\theta)=-\tan \theta; \tan (90^{0}+\theta)=-\cot \theta]$
= $\frac{ -\lambda+1/\lambda}{1+1}$
=$\frac{-\lambda^{2}+1}{2 \lambda}$
=$\frac{1- \lambda^{2}}{2 \lambda}$
