Question

If tan3θ = cotθ, then θ =

Options

• \[\frac{\left(2\mathrm{n}+1\right)\pi}{8},\mathrm{n}\in\mathbb{Z}\]

• \[\frac{(2\mathrm{n}+1)\pi}{4},\mathrm{n}\in\mathbb{Z}\]

• \[\frac{(n+2)\pi}{3},n\in\mathbb{Z}\]

• nπ, n ε Ζ

Answer 1

\[\frac{\left(2\mathrm{n}+1\right)\pi}{8},\mathrm{n}\in\mathbb{Z}\]

Explanation:

tan3θ = cotθ 

\[\therefore\quad\tan3\theta=\tan\left(\frac{\pi}{2}-\theta\right)\]

\[\therefore\quad3\theta=\mathrm{n}\pi+\frac{\pi}{2}+\theta,\mathrm{n}\in\mathbb{Z}\]

\[\therefore\quad4\theta=\mathrm{n}\pi+\frac{\pi}{2}\]

\[\therefore\quad\theta=\frac{\left(2\mathrm{n}+1\right)\pi}{8},\mathrm{n}\in\mathbb{Z}\]