Question

If $\tan {20}^{\circ }=k,$ Show that $\frac{\tan {160}^{\circ }-\tan {110}^{\circ }}{1+\tan {160}^{\circ }\tan {110}^{\circ }}=\frac{1-k^2}{2k}$

Answer 1

Consider, R.H.S. $=\frac{\tan {160}^{\circ }-\tan {110}^{\circ }}{1+\tan {160}^{\circ }\tan {110}^{\circ }}$

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

$=\frac{-\tan {20}^{\circ }+\cot {20}^{\circ }}{1-(-\tan {20}^{\circ })(-\cot {20}^{\circ })}$

$=\frac{-k+\frac{1}{k}}{1+1}$

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

Hence proved.