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

  • question_answer
    If \[{{\tan }^{2}}\theta -(1+\sqrt{3})\tan \theta +\sqrt{3}=0\], then the general value of \[\theta \] is

    A) \[n\pi +\frac{\pi }{4},n\pi +\frac{\pi }{3}\]

    B) \[n\pi -\frac{\pi }{4},n\pi +\frac{\pi }{3}\]

    C) \[n\pi +\frac{\pi }{4},n\pi -\frac{\pi }{3}\]

    D) \[n\pi -\frac{\pi }{4},n\pi -\frac{\pi }{3}\]

    Correct Answer: A

    Solution :

    \[{{\tan }^{2}}\theta -\tan \theta -\sqrt{3}\tan \theta +\sqrt{3}=0\] \[\Rightarrow \] \[\tan \theta (\tan \theta -1)-\sqrt{3}(\tan \theta -1)=0\] \[\Rightarrow \]\[(\tan \theta -\sqrt{3})\,\,(\tan \theta -1)=0\]\[\Rightarrow \]\[\theta =n\pi +\frac{\pi }{3}\], \[n\pi +\frac{\pi }{4}\].


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