Assume that ,
x2=cos2θ;θ=21cos−1x2
∴tan−1[1+cos2θ−1−cos2θ1+cos2θ+1−cos2θ]=tan−1[2cos2θ−2sin2θ2cos2θ+2sin2θ]=tan−1[cosθ−sinθcosθ+sinθ]
=tan−1[1−tanθ1+tanθ]
=tan−1[tan(4π+θ)]
=4π+21cos−1x2
The value of tan−1[1+x2−1−x21+x2+1−x2], ∣x∣<21,x=0, is equal to:
Held on 8 Apr 2017 · Verified 6 Jul 2026.
4π+21cos−1x2
4π−cos−1x2
4π−21cos−1x2
4π+cos−1x2
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