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Question: General solution of \[\tan \theta + \tan 4\theta + \tan 7\theta = \tan \theta \tan 4\theta \tan 7\th...

General solution of tan⁡θ+tan⁡4θ+tan⁡7θ=tan⁡θtan⁡4θtan⁡7θ\tan \theta + \tan 4\theta + \tan 7\theta = \tan \theta \tan 4\theta \tan 7\theta is
a. θ=nπ12\theta = n\dfrac{\pi }{{12}} where, n∈Zn \in \mathbb{Z}
b. θ=nπ9\theta = n\dfrac{\pi }{9} where, n∈Zn \in \mathbb{Z}
c. θ=nπ+π12\theta = n\pi + \dfrac{\pi }{{12}} where, n∈Zn \in \mathbb{Z}
d. none of these

Explanation

Solution

To deal with this problem we will use the help of the general formula of
  tan⁡A+tan⁡B1−tan⁡Atan⁡B=tan⁡(A+B)\;\dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} = \tan \left( {A + B} \right). We will then use the case of x=y+nπx = y + n\pi for, tan⁡x=tan⁡y\tan x = \tan y. Then analyzing the options, we will get the needed answer.

Complete step-by-step answer:
We have,
tanθ+tan4θ+tan7θ=tanθtan4θtan7θtan\theta + tan4\theta + tan7\theta = tan\theta tan4\theta tan7\theta
Changing sides, we get,
⇒tanθ+tan4θ=tanθtan4θtan7θ−tan⁡7θ\Rightarrow tan\theta + tan4\theta = tan\theta tan4\theta tan7\theta - \tan 7\theta
Taking −tan7θ- tan7\theta common we get,
⇒tanθ+tan4θ=−tan7θ(1−tanθtan4θ)\Rightarrow tan\theta + tan4\theta = - tan7\theta \left( {1 - tan\theta tan4\theta } \right)
Dividing both sides with, 1−tanθtan4θ1 - tan\theta tan4\theta we get,
⇒tanθ+tan4θ1−tanθtan4θ=−tan7θ\Rightarrow \dfrac{{tan\theta + tan4\theta }}{{1 - tan\theta tan4\theta }} = - tan7\theta
Since,   tanA+tanB1−tanAtanB=tan(A+B)\;\dfrac{{tanA + tanB}}{{1 - tanAtanB}} = tan\left( {A + B} \right)
⇒tan(θ+4θ)=−tan7θ    \Rightarrow tan\left( {\theta + 4\theta } \right) = - tan7\theta \;\;
On Adding terms in bracket, we get,
⇒tan5θ=tan(−7θ)  \Rightarrow tan5\theta = tan\left( { - 7\theta } \right)\;
As, x=y+nπx = y + n\pi for, tan⁡x=tan⁡y\tan x = \tan y, we get,
⇒5θ=(−7θ)  +nπ\Rightarrow 5\theta = \left( { - 7\theta } \right)\; + n\pi
On simplification we get,
⇒12θ=nπ\Rightarrow 12\theta = n\pi
∴θ=nπ12,        ∀  n∈Z  \therefore \theta = \dfrac{{n\pi }}{{12}},\;{\text{ }}\;{\text{ }}\;\forall \;n \in \mathbb{Z}\;
Hence, option (a) is correct.

Note: We have,
tan θ = tan∝tan{\text{ }}\theta {\text{ }} = {\text{ }}tan \propto
Using tan⁡θ=sin θcos⁡θ\tan \theta = \dfrac{{sin{\text{ }}\theta }}{{\cos \theta }} , we get,
⇒sin θcos⁡θ − sin∝cos∝= 0\Rightarrow \dfrac{{sin{\text{ }}\theta }}{{\cos \theta }}{\text{ }} - {\text{ }}\dfrac{{sin \propto }}{{cos \propto }} = {\text{ }}0
On taking LCM we get,
⇒(sin θ cos∝− cos θ sin∝)cos θ cos∝= 0\Rightarrow \dfrac{{(sin{\text{ }}\theta {\text{ }}cos \propto - {\text{ }}cos{\text{ }}\theta {\text{ }}sin \propto )}}{{cos{\text{ }}\theta {\text{ }}cos \propto }} = {\text{ }}0
Using sin (θ −∝)=sin θ cos∝− cos θ sin∝sin{\text{ }}(\theta {\text{ }} - \propto ) = sin{\text{ }}\theta {\text{ }}cos \propto - {\text{ }}cos{\text{ }}\theta {\text{ }}sin \propto , we get,
⇒sin (θ −∝)cos θ cos∝= 0\Rightarrow \dfrac{{sin{\text{ }}(\theta {\text{ }} - \propto )}}{{cos{\text{ }}\theta {\text{ }}cos \propto }} = {\text{ }}0
On simplification we get,
⇒sin (θ −∝) = 0\Rightarrow sin{\text{ }}(\theta {\text{ }} - \propto ){\text{ }} = {\text{ }}0
⇒ (θ − ∝) = nπ,\Rightarrow {\text{ }}\left( {\theta {\text{ }} - {\text{ }} \propto } \right){\text{ }} = {\text{ }}n\pi ,where n∈Z (i.e., n = 0, ± 1, ± 2, ± 3,…….),n \in Z{\text{ }}\left( {i.e.,{\text{ }}n{\text{ }} = {\text{ }}0,{\text{ }} \pm {\text{ }}1,{\text{ }} \pm {\text{ }}2,{\text{ }} \pm {\text{ }}3, \ldots \ldots .} \right), [Since we know that the θ = nπ, n∈Z\theta {\text{ }} = {\text{ }}n\pi ,{\text{ }}n \in Z is the general solution of the given equation sin θ = 0sin{\text{ }}\theta {\text{ }} = {\text{ }}0]
⇒θ = nπ +∝,\Rightarrow \theta {\text{ }} = {\text{ }}n\pi {\text{ }} + \propto , where n∈Z (i.e., n = 0, ± 1, ± 2, ± 3,…….),n \in Z{\text{ }}\left( {i.e.,{\text{ }}n{\text{ }} = {\text{ }}0,{\text{ }} \pm {\text{ }}1,{\text{ }} \pm {\text{ }}2,{\text{ }} \pm {\text{ }}3, \ldots \ldots .} \right),