Q73 / 160 · AP EAMCET 2022 (06 Jul Shift 2) · Mathematics
Let $P$ be a real number and $|P| \geq 2$. If $A, B, C$ are variable angles such that
$$
\left(\sqrt{P^2-4}\right) \tan A+P \tan B+\left(\sqrt{P^2+4}\right) \tan C=6 P \text {, }
$$
then the minimum value of
$$
\tan ^2 A+\tan ^2 B+\tan ^2 C=
$$
- A6
- B8
- Correct answerC12
- D18
Solution
Let $\mathbf{a}=\sqrt{P^2-4} \hat{i}+P \hat{j}+\sqrt{P^2+4} \hat{k}$ and $\mathbf{b}=\tan A \hat{i}+\tan B \hat{j}+\tan C \hat{k}$ are two vector expression
$\mathbf{a} \cdot \mathbf{b}=\sqrt{P^2-4} \tan A+P \tan B+\sqrt{P^2+4} \tan C$ $|\mathbf{a}||\mathbf{b}| \cos \theta=6 P$, where $Q$ is the angle between $\mathbf{a}$ and $\mathbf{b}$
$$
\begin{aligned}
& \sqrt{P^2-4+P^2+P^2+4} \sqrt{\tan ^2 A+\tan ^2 B+\tan ^2 C} \\
& \cos \theta=6 P
\end{aligned}
$$
$$
\begin{aligned}
& \cos \theta=6 P \\
& \sqrt{3} P \sqrt{\tan ^2 A+\tan ^2 B+\tan ^2 C}=6 P \sec \theta
\end{aligned}
$$
Squaring both sides,
$$
\begin{aligned}
& \text { Squaring both sides, } \\
& 3\left[\tan ^2 A+\tan ^2 B+\tan ^2 C\right]=36 \sec ^2 \theta
\end{aligned}
$$
$$
\Rightarrow \tan ^2 A+\tan ^2 B+\tan ^2 C=12 \sec ^2 \theta
$$
As we know that $\sec ^2 \theta \geq 1,12 \sec ^2 \theta \geq 12$
$$
\tan ^2 A+\tan ^2 B+\tan ^2 C \geq 12
$$
$\therefore$ Minimum value of $\tan ^2 A+\tan ^2 B+\tan ^2 C=12$
Hence, the correct answer is option: 'C'.