Question

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

Answer 1

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