If x∈(1,∞), then sin−12x1+x2 equals
2 tan−1x
π -2 tan−1x
-π -2 tan−1x
none of these
Let tan−1x=θ. Then, x=tanθ
Now,
x∈(1,∞)⇒tanθ>1⇒π4<θ<π2⇒π2<2θ<π
∴ sin−12x1+x2 = sin−1(sin2θ) = sin−1(sin(π−2θ))
= π−2θ ∵π4<θ<π2⇒−π2<π−2θ<0
=π−2tan−1x