Mathematics | TS EAMCET 2017 | Discussion Page

If $\tan 20^{0}= \lambda ,$ then

$\frac{\tan 160^{0}-\tan 110^{0}}{1+(\tan 160^{0})(\tan 110^{0})}=$

$\frac{1+\lambda^{2}}{2\lambda}$

$\frac{1+\lambda^{2}}{\lambda}$

$\frac{1-\lambda^{2}}{\lambda}$

$\frac{1-\lambda^{2}}{2\lambda}$

Option D

Given, $\tan 20^{0}= \lambda$

$\therefore$ $\frac{\tan 160^{0}-\tan 110^{0}}{1+ (\tan 160^{0})(\tan 110^{0}) }$

= $\frac{\tan (180^{0}-20^{0})-\tan (90^{0}+20^{0})}{1+ (\tan( 180^{0}-20^{0})(\tan 90^{0}+20^{0}) }$

$=\frac{-\tan 20^{0}+\cot 20^{0}}{1+ \tan 20^{0} \cot 20^{0}}$

$[ \because \tan (180^{0}-\theta)=-\tan \theta; \tan (90^{0}+\theta)=-\cot \theta]$

= $\frac{ -\lambda+1/\lambda}{1+1}$

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

= $\frac{1- \lambda^{2}}{2 \lambda}$

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