The general solution of tan⁡θ+tan⁡4θ+tan⁡7θ=tan⁡θ⋅tan⁡4θ⋅tan⁡7θ

We have tan⁡θ+tan⁡4θ+tan⁡7θ=tan⁡θ⋅tan⁡4θ⋅tan⁡7θ⇒ tan⁡θ+tan⁡4θ=−tan⁡7θ+tan⁡θ⋅tan⁡4θ⋅tan⁡7θ⇒ tan⁡θ+tan⁡4θ=−tan⁡7θ(1−tan⁡θ⋅tan⁡4θ)⇒tan⁡θ+tan⁡4θ1−tan⁡θ⋅tan⁡4θ=−tan⁡7θ⇒ tan⁡(θ+4θ)=−tan1⁡7θ⇒ tan⁡5θ=tan⁡(−7θ)⇒ 5θ=nπ+(−7θ)⇒ 12θ=nπ⇒θ=nπ12 (where n∈I )

The general solution of tan⁡θ+tan⁡4θ+tan⁡7θ=tan⁡θ⋅tan⁡4θ⋅tan⁡7θ

We have tan⁡θ+tan⁡4θ+tan⁡7θ=tan⁡θ⋅tan⁡4θ⋅tan⁡7θ⇒ tan⁡θ+tan⁡4θ=−tan⁡7θ+tan⁡θ⋅tan⁡4θ⋅tan⁡7θ⇒ tan⁡θ+tan⁡4θ=−tan⁡7θ(1−tan⁡θ⋅tan⁡4θ)⇒tan⁡θ+tan⁡4θ1−tan⁡θ⋅tan⁡4θ=−tan⁡7θ⇒ tan⁡(θ+4θ)=−tan1⁡7θ⇒ tan⁡5θ=tan⁡(−7θ)⇒ 5θ=nπ+(−7θ)⇒ 12θ=nπ⇒θ=nπ12 (where n∈I )

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The general solution of $\tan θ + \tan 4 θ + \tan 7 θ = \tan θ \cdot \tan 4 θ \cdot \tan 7 θ$

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$θ = \frac{n π}{4}$

$θ = \frac{n π}{12}$

$θ = \frac{n π}{8}$

None of these

answer is B.

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Detailed Solution

$\begin{array}{l} We have \tan θ + \tan 4 θ + \tan 7 θ = \tan θ \cdot \tan 4 θ \cdot \tan 7 θ \\ \Rightarrow \tan θ + \tan 4 θ = - \tan 7 θ + \tan θ \cdot \tan 4 θ \cdot \tan 7 θ \\ \Rightarrow \tan θ + \tan 4 θ = - \tan 7 θ (1 - \tan θ \cdot \tan 4 θ) \\ \Rightarrow \frac{\tan θ + \tan 4 θ}{1 - \tan θ \cdot \tan 4 θ} = - \tan 7 θ \\ \Rightarrow \tan (θ + 4 θ) = - \tan_{1} 7 θ \\ \Rightarrow \tan 5 θ = \tan (- 7 θ) \\ \Rightarrow 5 θ = n π + (- 7 θ) \\ \Rightarrow 12 θ = n π \Rightarrow θ = \frac{n π}{12} (where n \in I) \end{array}$