Since tan−1(1−x22x)=2tan−1x:
tan−1y=3tan−1x, so y=tan(3θ) where θ=tan−1x.
y=1−3tan2θ3tanθ−tan3θ=1−3x23x−x3.
Let tan−1y=tan−1x+tan−1(1−x22x). Then y is :
Held on 15 Jun 2023 · Verified 13 Jul 2026.
1−3x23x−x3
1−3x23x+x3
1+3x23x−x3
1+3x23x+x3
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