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प्रश्न
If tan3θ = cotθ, then θ =
विकल्प
\[\frac{\left(2\mathrm{n}+1\right)\pi}{8},\mathrm{n}\in\mathbb{Z}\]
\[\frac{(2\mathrm{n}+1)\pi}{4},\mathrm{n}\in\mathbb{Z}\]
\[\frac{(n+2)\pi}{3},n\in\mathbb{Z}\]
nπ, n ε Ζ
MCQ
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उत्तर
\[\frac{\left(2\mathrm{n}+1\right)\pi}{8},\mathrm{n}\in\mathbb{Z}\]
Explanation:
tan3θ = cotθ
\[\therefore\quad\tan3\theta=\tan\left(\frac{\pi}{2}-\theta\right)\]
\[\therefore\quad3\theta=\mathrm{n}\pi+\frac{\pi}{2}+\theta,\mathrm{n}\in\mathbb{Z}\]
\[\therefore\quad4\theta=\mathrm{n}\pi+\frac{\pi}{2}\]
\[\therefore\quad\theta=\frac{\left(2\mathrm{n}+1\right)\pi}{8},\mathrm{n}\in\mathbb{Z}\]
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