Question

In any $\Delta ABC$, if $y=\tan A\tan B+\tan B\tan C+\tan C\tan A$, then

• A. $y\in (0,9)$
• B. $y\notin (0,9)$
• C. $y\in [0,9]$
• D. None of these

Answer 1

The correct option is B $y\notin (0,9)$

In any $△ABC$
$A+B=\pi -C\Rightarrow \tan (A+B)=\tan (\pi -C)\Rightarrow \frac{\tan A+\tan B}{1-\tan A\tan B}=-\tan C$
$\Rightarrow \tan A+\tan B+\tan C=\tan A\tan B\tan C$ ...(1)
$y=\tan A\tan B+\tan B\tan C+\tan A\tan C$
$\Rightarrow y=\tan A\tan B\tan C(\frac{1}{\tan A}+\frac{1}{\tan B}+\frac{1}{\tan C})$
$\Rightarrow y=(\tan A+\tan B+\tan C)(\frac{1}{\tan A}+\frac{1}{\tan B}+\frac{1}{\tan C})$ ...{ From 1}

$\because A.M\ge H.M$
$\therefore \frac{\tan A+\tan B+\tan C}{3}\ge \frac{3}{\frac{1}{\tan A}+\frac{1}{\tan B}+\frac{1}{\tan C}}$
$\Rightarrow (\tan A+\tan B+\tan C)(\frac{1}{\tan A}+\frac{1}{\tan B}+\frac{1}{\tan C})\ge 9\Rightarrow y\ge 9$
Ans: B