The general solution of the equation θ · 2 θ=1 is...

Correct Answer is : θ=nπ±π6,n∈Zθ=n π ± π/6, n Zθ=nπ±6π,n∈Z

Correct Answer is : θ=nπ±π6,n∈Zθ=n π ± π/6, n Zθ=nπ±6π,n∈Z

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MHTCET 16 Apr 2026 Shift 2 Paper

Question : 104 of 150

Marks: +1, -0

\cot \theta \cdot \cot 2 \theta=1

is...

Solution:

Concept:

\cot 2\theta

can be expressed in terms of

\cot \theta

using the double-angle identity, then the resulting trigonometric equation is solved for its general solution.

Explanation:

Start with

\cot \theta \cdot \cot 2\theta = 1

Use the identity

\cot 2\theta = \frac{\cot^2 \theta - 1}{2\cot \theta}

Substitute into the equation:

\cot \theta \cdot \frac{\cot^2 \theta - 1}{2\cot \theta} = 1

Cancel

\cot \theta

(valid when

\cot \theta \neq 0

\frac{\cot^2 \theta - 1}{2} = 1

Solve step by step:

\cot^2 \theta - 1 = 2

\cot^2 \theta = 3

Hence,

\cot \theta = \pm \sqrt{3}

Since

\cot \frac{\pi}{6} = \sqrt{3}

and

\cot\left(-\frac{\pi}{6}\right) = -\sqrt{3}

, the principal solutions are

\pm \frac{\pi}{6}

The general solution for

\cot \theta = \pm \sqrt{3}

\theta = n\pi \pm \frac{\pi}{6}, \quad n \in \mathbb{Z}

Answer:

Option A:

\theta = n\pi \pm \frac{\pi}{6}, \ n \in \mathbb{Z}

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