If $\tan \,\,{\rm{\theta }} + \tan \,\,2{\rm{\theta }} + \sqrt 3 \,\,\tan \,\,{\rm{\theta }}\,\,\tan \,\,2{\rm{\theta }} = \sqrt 3 $, then

${\rm{\theta }} = \frac{{\left( {6\,n + 1} \right)\pi }}{{18}}\,,\,\,\forall \;n \in I$

${\rm{\theta }} = \frac{{\left( {6\,n + 1} \right)\pi }}{9}\,,\,\,\forall \;n \in I$

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

\tan \,\,\mathrm{\theta } + \tan \,\,2\mathrm{\theta } + \sqrt 3 \,\,\tan \,\,\mathrm{\theta }\,\,\tan \,\,2\mathrm{\theta } = \sqrt 3

, then

View answer and explanation

Correct option: C

\mathrm{\theta } = \frac{{\left( {3\,n + 1} \right)\pi }}{9},\,\,\forall \;n \in I

Option D (

\mathrm{\theta } = \frac{{(3n + 1)\pi }}{9},\\,\forall \;n \in I

) is correct because it is derived from solving the trigonometric equation using the tangent addition formula.

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