The values of sec2⁡tan−1⁡2+cosec2⁡cot−1⁡3=

The values of sec2tan12+cosec2cot13=

  1. A

    5

  2. B

    10

  3. C

    15

  4. D

    20

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    Solution:

    We have,

    sec2tan12+cosec2cot13=sectan122+coseccot132=sectan1212+coseccot1312=secsec152+coseccosec1102=(5)2+(10)2=15

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