Consider tan−1[sin(cos−132)] Let cos−132=θ⇒cosθ=32⇒sinθ=1−cos2θ=1−32=31∴tan−1[sin(cos−132)]=tan−1[sinθ]=tan−1[31]=tan−1(31)=6π
A value of tan−1(sin(cos−1(32))) is
Held on 19 May 2012 · Verified 6 Jul 2026.
4π
2π
3π
6π
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