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

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

    A) \[n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{3}\]

    B) \[n\pi +{{(-1)}^{n}}\frac{\pi }{3}-\frac{\pi }{4}\]

    C) \[n\pi +{{(-1)}^{n}}\frac{\pi }{3}+\frac{\pi }{4}\]

    D) \[n\pi +{{(-1)}^{n}}\frac{\pi }{4}+\frac{\pi }{3}\]

    Correct Answer: B

    Solution :

    \[\sin \theta +\cos \theta =\sqrt{2}\cos \alpha \Rightarrow \cos \left( \theta -\frac{\pi }{4} \right)=\cos \alpha \] \[\Rightarrow \] \[\theta -\frac{\pi }{4}=2n\pi \pm \alpha \Rightarrow \theta =2n\pi +\frac{\pi }{4}\pm \alpha \].


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