Q41 / 160 · TS EAMCET 2017 · Mathematics
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$\frac{1-\lambda^2}{\lambda}$
- Correct answerD$\frac{1-\lambda^2}{2 \lambda}$
Solution
Given, $\tan 20^{\circ}=\lambda$
$$
\begin{aligned}
\therefore & \frac{\tan 160^{\circ}-\tan 110^{\circ}}{1+\left(\tan 160^{\circ}\right)\left(\tan 110^{\circ}\right)} \\
& =\frac{\tan \left(180^{\circ}-20^{\circ}\right)-\tan \left(90^{\circ}+20^{\circ}\right)}{1+\left(\tan \left(180^{\circ}-20^{\circ}\right)\left(\tan \left(90^{\circ}+20^{\circ}\right)\right)\right.} \\
& =\frac{-\tan 20^{\circ}+\cot 20^{\circ}}{1+\tan 20^{\circ} \cot 20^{\circ}} \\
& {\left[\because \tan \left(180^{\circ}-\theta\right)=-\tan \theta ; \tan \left(90^{\circ}+\theta\right)=-\cot \theta\right] } \\
= & \frac{-\lambda+1 / \lambda}{1+1} \\
= & \frac{-\lambda^2+1}{2 \lambda} \\
= & \frac{1-\lambda^2}{2 \lambda}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.