JEE Main & Advanced Sample Paper JEE Main Sample Paper-22

  • question_answer
    If\[\frac{\sqrt{1+\sin \frac{39\pi }{8}}}{\sqrt{1+\sin \frac{57\pi }{8}}}=\tan \left( \frac{k\pi }{16} \right)\]   then least positive value of k is

    A)  1                                

    B)  3

    C)  5                                

    D)  7

    Correct Answer: C

    Solution :

    \[\frac{\sqrt{1+\sin \,\frac{39\pi }{8}}}{\sqrt{1+\sin \,\frac{57\pi }{8}}}\,=\frac{\sqrt{1+\sin \frac{\pi }{8}}}{\sqrt{1-\sin \frac{\pi }{8}}}\]           \[=\frac{\sqrt{1+\cos \frac{3\pi }{8}}}{\sqrt{1-\cos \frac{3\pi }{8}}}\,=\frac{\sqrt{2}\cos \,\frac{3\pi }{16}}{\sqrt{2-\sin \frac{3\pi }{16}}}\]\[=\cot \,\frac{3\pi }{16}=\tan \,=\frac{5\pi }{16}\]             \[\therefore \] least positive value of k is 5.


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