$$tan$$−$$1$$⁡$$($$$$tan$$⁡$$1$$−θ$$)=1$$−θ when

Correct option is 44−π24<θ≤1

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$\tan^{- 1} (\tan \sqrt{1 - θ}) = \sqrt{1 - θ}$ when

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−π2<θ<π2

θ>4−π4

θ<4−π4

4−π24<θ≤1

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detailed solution

Correct option is D

Clearly, 1−θ is real, If θ≤1.Now,tan−1⁡(tan⁡1−θ)=1−θ⇒ 0≤1−θ<π2 and θ≤1⇒ 0≤1−θ<π24 and θ≤1⇒ −1≤−θ≤π24−1 and θ≤1⇒ 1−π24<θ≤1 and θ≤1⇒4−π24<θ≤1

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tan−1⁡(tan⁡1−θ)=1−θ when

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$\tan^{- 1} (\tan \sqrt{1 - θ}) = \sqrt{1 - θ}$ when