Question

If $k$ is a real constant and $A,B,C$ are variable angles such that $\left(\sqrt{k^2-4}\right)\tan A+k\tan B+\left(\sqrt{k^2+4}\right)\tan C=6k,$ then the minimum value of ${\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}C$ is

• A. $1$
• B. $9$
• C. $\sqrt{5}$
• D. $12$

Answer 1

The correct option is D $12$

Let the two vectors be,
$\overrightarrow{a}=\left(\sqrt{k^2-4}\right)\hat{i}+k\hat{j}+\left(\sqrt{k^2+4}\right)\hat{k}$
and $\overrightarrow{b}=\left(\tan A\right)\hat{i}+\left(\tan B\right)\hat{j}+\left(\tan C\right)\hat{k}$
Given, $\overrightarrow{a}\cdot \overrightarrow{b}=6k$
$\Rightarrow |\overrightarrow{a}||\overrightarrow{b}|\cos \theta =6k\Rightarrow \sqrt{3k^2}\sqrt{{\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}C}=6k\sec \theta$
Squaring both the sides,
${\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}C=12{\sec }^2\theta$
Since ${\sec }^2\theta \ge 1,$
hence, the minimum value is $12$.