Question

If $A={110}^{\circ },$ then show that $\frac{1+\sqrt{1+{\tan }^{2}2A}}{\tan 2A}=-\tan A$.

Answer 1

$\frac{1+\sqrt{1+{\tan }^{2}2A}}{\tan 2A}=\frac{1+\sqrt{1+{\tan }^2\left({220}^0\right)}}{\tan \left({220}^0\right)}$
$=\frac{1+\sqrt{1+{\tan }^2({180}^0+{40}^0)}}{\tan ({180}^0+{40}^0)}=\frac{1+\sqrt{1+{\tan }^2{40}^0}}{\tan {40}^0}$
$=\frac{1+\sec {40}^0}{\tan {40}^0}=\frac{1+\frac{1}{\cos {40}^0}}{\frac{\sin {40}^0}{\cos {40}^0}}=\frac{\cos {40}^0+1}{\sin {40}^0}$
$=\frac{2{\cos }^2{20}^0-1+1}{2\sin {20}^0\cos {20}^0}=\cot {20}^0=-\tan {110}^0=-\tan A$