The value of x where x >0 and tan⁡sec−1⁡1x=sin⁡tan−1⁡2, is 

The value of x where x >0 and tansec11x=sintan12, is 

  1. A

    5

  2. B

    53

  3. C

    1

  4. D

    23

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    Solution:

    We have, 

    tansec11x=sintan12tantan11x2x=sinsin1251x2x=251x21=451x2=95x=53

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