Question

If $\tan \, 20^{\circ} = \lambda$ , then $\frac{\tan160^{\circ} - \tan110^{\circ}}{1+\left(\tan160^{\circ}\right) \left(\tan110^{\circ}\right) }$ =

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