Question

If $A={110}^o$ then $\frac{1+\sqrt{1+{\tan }^{2}2A}}{\tan 2A}$ equal ?

• A. $\tan A$
• B. $-\tan A$
• C. $\cot A$
• D. $-\cot A$

Answer 1

The correct option is B $-\tan A$

Given $A={110}^{\circ }⟹\text{sec}2A<0$

So $\frac{1+\sqrt{1+{\tan }^{2}2A}}{\tan 2A}=\frac{1+\sqrt{{\text{sec}}^{2}2A}}{\tan 2A}$

$=\frac{1+\mid \text{sec}2A\mid }{\tan 2A}$

$=\frac{1-\text{sec}2A}{\tan 2A}$ $(\because \text{sec}2A<0)$

$=\frac{1-\frac{1}{\cos 2A}}{\frac{\sin 2A}{\cos 2A}}$

$=\frac{\cos 2A-1}{\sin 2A}$

$=\frac{-2{\sin }^{2}A}{2\sin A\cos A}$

$=\frac{-\sin A}{\cos A}$

$=-\tan A$