Question

If $A,B,C$ are the angles of triangle such that $0<A\le \frac{\pi }{3}$, then the range of $\frac{\tan B+\tan C}{\tan B\tan C-1}$ is

• A. $(-\sqrt{3},0)$
• B. $(0,\sqrt{3}]$
• C. $(0,\sqrt{3})$
• D. $[0,\sqrt{3}]$

Answer 1

The correct option is B $(0,\sqrt{3}]$

Given:
$\frac{\tan B+\tan C}{\tan B\tan C-1}=-\left(\frac{\tan B+\tan C}{1-\tan B\tan C}\right)=-\tan (B+C)=-\tan (\pi -A)(\because A+B+C=\pi )=\tan A$

It is given that,
$0<A\le \frac{\pi }{3}$
$\Rightarrow 0<\tan A\le \sqrt{3}$
Therefore,
$0<\frac{\tan B+\tan C}{\tan B\tan C-1}\le \sqrt{3}$