The correct option is B √x2−1
Given: θ is an acute angle and sinθ2=√x−12x
To find: tanθ
Now, from the sub multiple angle formulae of tanθ
i.e. tanθ=2tanθ21−tan2θ2
Now, 0<θ<π2⇒0<θ2<π4
Hence, all trigonometric ratios will be positive.
Now, using the identity sin2A+cos2A=1
we can say, cosθ2=√1−sin2θ2
⇒cosθ2=√1−x−12x
⇒cosθ2=√x+12x
⇒tanθ2=sinθ2cosθ2=√x−1x+1
Substituting in the formulae for submultiple angle, we get:
tanθ=2√x−1x+11−x−1x+1
⇒tanθ=2√x−1x+12x+1
⇒tanθ=√x−1x+1×(x+1)2=√x2−1
∴tanθ=√x2−1