Question

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

• A. $\frac{1+{\lambda }^2}{2\lambda }$
• B. $\frac{1+{\lambda }^2}{\lambda }$
• C. $\frac{1-{\lambda }^2}{\lambda }$
• D. $\frac{1-{\lambda }^2}{2\lambda }$

Answer 1

Answer: (D)

Given, $\tan 20^{\circ }=\lambda$
$\therefore \frac{\tan 160^{\circ }-\tan 110^{\circ }}{1+\left(\tan 160^{\circ }\right)\left(\tan 110^{\circ }\right)}$
$=\frac{\tan \left(180^{\circ }-20^{\circ }\right)-\tan \left(90^{\circ }+20^{\circ }\right)}{1+(\tan \left(180^{\circ }-20^{\circ }\right)\left(\tan \left(90^{\circ }+20^{\circ }\right)\right)}$
$=\frac{-\tan 20^{\circ }+\cot 20^{\circ }}{1+\tan 20^{\circ }\cot 20^{\circ }}$
$\left[\because \tan \left(180^{\circ }-\theta \right)=-\tan \theta ;\tan \left(90^{\circ }+\theta \right)=-\cot \theta \right]$
$=\frac{-\lambda +1/\lambda }{1+1}$
$=\frac{-{\lambda }^2+1}{2\lambda }$
$=\frac{1-{\lambda }^2}{2\lambda }$