Q65 / 150 · MHT CET 2026 (16 April Shift 2) · Mathematics
The general solution of the equation $\cot\theta \cdot \cot 2\theta = 1$ is...
- Correct answerA$\theta = n\pi \pm \dfrac{\pi}{6}, n \in Z$
- B$\theta = n\pi \pm \dfrac{\pi}{3}, n \in Z$
- C$\theta = n\pi \pm \dfrac{\pi}{4}, n \in Z$
- D$\theta = n\pi \pm \dfrac{\pi}{8}, n \in Z$
Solution
Given equation is $\cot\theta \cdot \cot 2\theta = 1$
Writing in terms of $\tan\theta$:
$\dfrac{1}{\tan\theta} \cdot \dfrac{1 - \tan^2\theta}{2\tan\theta} = 1$
$1 - \tan^2\theta = 2\tan^2\theta$
$3\tan^2\theta = 1$
$\tan^2\theta = \dfrac{1}{3}$
$\tan^2\theta = \tan^2\left(\dfrac{\pi}{6}\right)$
Using the general solution for $\tan^2\theta = \tan^2\alpha$, which is $\theta = n\pi \pm \alpha, n \in Z$, we get:
$\theta = n\pi \pm \dfrac{\pi}{6}, n \in Z$
Answer: $\theta = n\pi \pm \dfrac{\pi}{6}, n \in Z$
Writing in terms of $\tan\theta$:
$\dfrac{1}{\tan\theta} \cdot \dfrac{1 - \tan^2\theta}{2\tan\theta} = 1$
$1 - \tan^2\theta = 2\tan^2\theta$
$3\tan^2\theta = 1$
$\tan^2\theta = \dfrac{1}{3}$
$\tan^2\theta = \tan^2\left(\dfrac{\pi}{6}\right)$
Using the general solution for $\tan^2\theta = \tan^2\alpha$, which is $\theta = n\pi \pm \alpha, n \in Z$, we get:
$\theta = n\pi \pm \dfrac{\pi}{6}, n \in Z$
Answer: $\theta = n\pi \pm \dfrac{\pi}{6}, n \in Z$
Hence, the correct answer is option: 'A'.