If tan⁡θ+tan⁡2θ+3tan⁡θtan⁡2θ=3,then

A
$θ = \frac{(6 n + 1) π}{18}, \forall n \in I$
B
$θ = \frac{(6 n + 1) π}{9}, \forall n \in I$
C
$θ = \frac{(3 n + 1) π}{9}, \forall n \in I$
D
None of these

Solution:

$\begin{aligned} ∵ \tan θ + \tan 2 θ + \sqrt{3} \tan θtan 2 θ = \sqrt{3} \\ \Rightarrow \tan θ + \tan 2 θ = \sqrt{3} (1 - \tan θtan 2 θ) \\ \Rightarrow \frac{\tan θ + \tan 2 θ}{1 - \tan θtan 2 θ} = \sqrt{3} \Rightarrow \tan 3 θ = \tan \frac{π}{3} \\ \Rightarrow 3 θ = nπ + \frac{π}{3} \Rightarrow θ = \frac{(3 n + 1) π}{9}, n \in I \end{aligned}$

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

Solution:

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