JEE Main & Advanced Mathematics Trigonometric Equations Question Bank Solution of trigonometrical equations

  • question_answer
    If \[\sqrt{3}\tan 2\theta +\sqrt{3}\tan 3\theta +\tan 2\theta \tan 3\theta =1\], then the general value of\[\theta \]is

    A) \[n\pi +\frac{\pi }{5}\]

    B) \[\left( n+\frac{1}{6} \right)\frac{\pi }{5}\]

    C) \[\left( 2n\pm \frac{1}{6} \right)\frac{\pi }{5}\]

    D) \[\left( n+\frac{1}{3} \right)\frac{\pi }{5}\]

    Correct Answer: B

    Solution :

    \[\sqrt{3}\tan 2\theta +\sqrt{3}\tan 3\theta +\tan 2\theta \tan 3\theta =1\] \[\Rightarrow \] \[\frac{\tan 2\theta +\tan 3\theta }{1-\tan 2\theta \tan 3\theta }=\frac{1}{\sqrt{3}}\] Þ \[\tan 5\theta =\tan \frac{\pi }{6}\] \[\Rightarrow \] \[5\theta =n\pi +\frac{\pi }{6}\Rightarrow \theta =\left( n+\frac{1}{6} \right)\frac{\pi }{5}\].


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