Question

General solution of \[\tan \theta + \tan 4\theta + \tan 7\theta = \tan \theta \tan 4\theta \tan 7\theta \] is
a. \[\theta = n\dfrac{\pi }{{12}}\] where, \[n \in \mathbb{Z}\]
b. \[\theta = n\dfrac{\pi }{9}\] where, \[n \in \mathbb{Z}\]
c. \[\theta = n\pi + \dfrac{\pi }{{12}}\] where, \[n \in \mathbb{Z}\]
d. none of these

Answer 1

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

Complete step-by-step answer:
We have,
 \[tan\theta + tan4\theta + tan7\theta = tan\theta tan4\theta tan7\theta \]
Changing sides, we get,
 \[ \Rightarrow tan\theta + tan4\theta = tan\theta tan4\theta tan7\theta - \tan 7\theta \]
Taking \[ - tan7\theta \] common we get,
 \[ \Rightarrow tan\theta + tan4\theta = - tan7\theta \left( {1 - tan\theta tan4\theta } \right)\]
Dividing both sides with, \[1 - tan\theta tan4\theta \] we get,
 \[ \Rightarrow \dfrac{{tan\theta + tan4\theta }}{{1 - tan\theta tan4\theta }} = - tan7\theta \]
Since, \[\;\dfrac{{tanA + tanB}}{{1 - tanAtanB}} = tan\left( {A + B} \right)\]
 \[ \Rightarrow tan\left( {\theta + 4\theta } \right) = - tan7\theta \;\;\]
On Adding terms in bracket, we get,
 \[ \Rightarrow tan5\theta = tan\left( { - 7\theta } \right)\;\]
As, \[x = y + n\pi \] for, \[\tan x = \tan y\], we get,
 \[ \Rightarrow 5\theta = \left( { - 7\theta } \right)\; + n\pi \]
On simplification we get,
 \[ \Rightarrow 12\theta = n\pi \]
 \[\therefore \theta = \dfrac{{n\pi }}{{12}},\;{\text{ }}\;{\text{ }}\;\forall \;n \in \mathbb{Z}\;\]
Hence, option (a) is correct.

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