tan−1⁡(tan⁡1−θ)=1−θ when 

tan1(tan1θ)=1θ when 

  1. A

    π2<θ<π2

  2. B

    θ>4π4

  3. C

    θ<4π4

  4. D

    4π24<θ1

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    Solution:

    Clearly,  1θ is real, If θ1.

    Now,

    tan1(tan1θ)=1θ

     01θ<π2 and θ1

     01θ<π24 and θ1

     1θπ241 and θ1

     1π24<θ1 and θ14π24<θ1

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