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Question

Mathematics Question on Trigonometric Functions

If tan θ \theta + tan 4θ4\theta + tan 7θ7\theta = tan θ\theta tan 4θ4\theta tan 7θ,7\theta, then the general solution is

A

θ=nπ4\theta=\frac{n\pi}{4}

B

θ=nπ12\theta=\frac{n\pi}{12}

C

θ=nπ6\theta=\frac{n\pi}{6}

D

none of these.

Answer

θ=nπ12\theta=\frac{n\pi}{12}

Explanation

Solution

tan⁡(θ+4θ+7θ)=(tan⁡θ+tan⁡4θ+tan⁡7θ−tan⁡θtan⁡4θtan⁡7θ)\tan (\theta + 4 \theta + 7\theta) = (\tan \theta + \tan 4 \theta + \tan 7 \theta - \tan \theta \tan 4 \theta \tan 7 \theta )
(1−tan⁡θtan⁡4θ−tan⁡4θtan⁡7θ−tan⁡7θtan⁡θ)=0(1 - \tan \theta \tan 4 \theta - \tan 4 \theta \tan 7 \theta - \tan 7 \theta \tan \theta) = 0
∵    tan⁡θ+tan⁡4θ+tan⁡7θ=tan⁡θtan⁡4θtan⁡7θ]\because \, \, \, \, \tan \theta + \tan 4 \theta + \tan 7 \theta = \tan \theta \tan 4 \theta \tan 7 \theta]
⇒   tan⁡(12θ)=0⇒12θ=nπ,n∈N\Rightarrow\, \, \, \tan (12\theta) = 0 \Rightarrow 12 \theta = n \pi , n \in N
⇒\Rightarrow θ=nπ12,n∈N ∴ (b)\theta = \frac{n \pi}{12} , n \in N \, \therefore \, (b) holds.