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Question: If \(\tan\theta + \tan 2\theta + \sqrt{3}\tan\theta\tan 2\theta = \sqrt{3},\) then...

If tan⁡θ+tan⁡2θ+3tan⁡θtan⁡2θ=3,\tan\theta + \tan 2\theta + \sqrt{3}\tan\theta\tan 2\theta = \sqrt{3}, then

A

θ=(6n+1)π18,∀n∈I\theta = \frac{(6n + 1)\pi}{18},\forall n \in I

B

θ=(6n+1)π9,∀n∈I\theta = \frac{(6n + 1)\pi}{9},\forall n \in I

C

θ=(3n+1)π9,∀n∈I\theta = \frac{(3n + 1)\pi}{9},\forall n \in I

D

None of these

Answer

θ=(3n+1)π9,∀n∈I\theta = \frac{(3n + 1)\pi}{9},\forall n \in I

Explanation

Solution

tan⁡θ+tan⁡2θ+3tan⁡θtan⁡2θ=3⇒tan⁡θ+tan⁡2θ=3(1−tan⁡θtan⁡2θ)⇒tan⁡θ+tan⁡2θ1−tan⁡θtan⁡2θ=3⇒tan⁡3θ=tan⁡(π3)⇒⥂⥂3θ=nπ+π3⇒θ=nπ3+π9=(3n+1)π9\tan\theta + \tan 2\theta + \sqrt{3}\tan\theta\tan 2\theta = \sqrt{3}\mathbf{\Rightarrow}\mathbf{\tan}\mathbf{\theta}\mathbf{+}\mathbf{\tan}\mathbf{2}\mathbf{\theta =}\sqrt{\mathbf{3}}\mathbf{(1 -}\mathbf{\tan}\mathbf{\theta}\mathbf{\tan}\mathbf{2}\mathbf{\theta) \Rightarrow}\frac{\mathbf{\tan}\mathbf{\theta}\mathbf{+}\mathbf{\tan}\mathbf{2}\mathbf{\theta}}{\mathbf{1 -}\mathbf{\tan}\mathbf{\theta}\mathbf{\tan}\mathbf{2}\mathbf{\theta}}\mathbf{=}\sqrt{\mathbf{3}}\mathbf{\Rightarrow}\mathbf{\tan}\mathbf{3}\mathbf{\theta =}\mathbf{\tan}\left( \frac{\mathbf{\pi}}{\mathbf{3}} \right)\mathbf{\Rightarrow ⥂ ⥂ 3\theta = n\pi +}\frac{\mathbf{\pi}}{\mathbf{3}}\mathbf{\Rightarrow}\mathbf{\theta =}\frac{\mathbf{n\pi}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{\pi}}{\mathbf{9}}\mathbf{= (3n + 1)}\frac{\mathbf{\pi}}{\mathbf{9}}