Question

In a $△ABC,\tan A\tan B\tan C=9$.For such triangles, if ${\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}C=\lambda$ then

• A. $9.3^{\frac{1}{3}}<\lambda <27$
• B. $\lambda$ is not lesser than or equal to $27$
• C. $\lambda <9.3^{\frac{1}{3}}$
• D. $\lambda$ is not lesser than $27$

Answer 1

The correct option is D $\lambda$ is not lesser than $27$

$\because {(\tan A+\tan B)}^2+{(\tan B+\tan C)}^2+{(\tan C+\tan A)}^2>0$
$\Rightarrow 2({\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}C-\tan A\tan B-\tan B\tan C-\tan C\tan A)>0$
$\Rightarrow 3({\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}C)-{(\tan A+\tan B+\tan C)}^2>0$
$3\lambda -{(\tan A+\tan B+\tan C)}^2>0$
$\Rightarrow 3\lambda -81>0$
$\Rightarrow \lambda >27$
Thus, $\lambda$ is not less then $27$