Question

If $\tan {20}^{\circ }=\lambda$ then show that $\frac{\tan {160}^{\circ }-\tan {110}^{\circ }}{1+\tan {160}^{\circ }\cdot \tan {110}^{\circ }}=\frac{1-{\lambda }^2}{2\lambda }$

Answer 1

Given $\tan {20}^{\circ }=\lambda$

$\tan {40}^{\circ }=\tan 2\left({20}^{\circ }\right)=\frac{2\tan {20}^{\circ }}{1-{\tan }^2{20}^{\circ }}=\frac{2\lambda }{1-{\lambda }^2}$

$LHS=\frac{\tan {160}^{\circ }-\tan {110}^{\circ }}{1+\tan {160}^{\circ }\tan {110}^{\circ }}=\tan ({160}^{\circ }-{110}^{\circ })=\tan {50}^{\circ }=\tan ({90}^{\circ }-{40}^{\circ })=\cot {40}^{\circ }=\frac{1}{\tan {40}^{\circ }}=\frac{1-{\lambda }^2}{2\lambda }=RHS$