Q92 / 150 · MHT CET 2026 (15 April Shift 2) · Mathematics
If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, $|\vec{a}| = |\vec{b}| = |\vec{c}| = 3$ and $\theta$ is the angle between $\vec{b}$ and $\vec{c}$ then $\tan^2\theta + \cot^2\theta =$
- A$\dfrac{2}{3}$
- B$\dfrac{5}{3}$
- C$\dfrac{8}{3}$
- Correct answerD$\dfrac{10}{3}$
Solution
Given $\vec{a} + \vec{b} + \vec{c} = \vec{0}$
$\vec{b} + \vec{c} = -\vec{a}$
Squaring both sides, we get
$|\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b}\cdot\vec{c} = |\vec{a}|^2$
Substituting $|\vec{a}| = |\vec{b}| = |\vec{c}| = 3$,
$3^2 + 3^2 + 2|\vec{b}||\vec{c}|\cos\theta = 3^2$
$9 + 9 + 2(3)(3)\cos\theta = 9$
$18 + 18\cos\theta = 9$
$18\cos\theta = -9 \Rightarrow \cos\theta = -\dfrac{1}{2}$
This gives $\theta = 120^\circ$
Now, $\tan^2\theta + \cot^2\theta = \tan^2(120^\circ) + \cot^2(120^\circ)$
$= (-\sqrt{3})^2 + \left(-\dfrac{1}{\sqrt{3}}\right)^2$
$= 3 + \dfrac{1}{3} = \dfrac{10}{3}$
Answer: $\dfrac{10}{3}$
$\vec{b} + \vec{c} = -\vec{a}$
Squaring both sides, we get
$|\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b}\cdot\vec{c} = |\vec{a}|^2$
Substituting $|\vec{a}| = |\vec{b}| = |\vec{c}| = 3$,
$3^2 + 3^2 + 2|\vec{b}||\vec{c}|\cos\theta = 3^2$
$9 + 9 + 2(3)(3)\cos\theta = 9$
$18 + 18\cos\theta = 9$
$18\cos\theta = -9 \Rightarrow \cos\theta = -\dfrac{1}{2}$
This gives $\theta = 120^\circ$
Now, $\tan^2\theta + \cot^2\theta = \tan^2(120^\circ) + \cot^2(120^\circ)$
$= (-\sqrt{3})^2 + \left(-\dfrac{1}{\sqrt{3}}\right)^2$
$= 3 + \dfrac{1}{3} = \dfrac{10}{3}$
Answer: $\dfrac{10}{3}$
Hence, the correct answer is option: 'D'.