The correct option is C 12
f(x)=(logcotxtan x)(logtan xcot x)+tan−1(x√(4−x2))
f(x)=1+tan−1(x√(4−x2))
Now Put x=2sinθ, we get
f(x)=1+tan−1(2sinθ√4−4sin2θ)⇒f(x)=1+tan−1(2sinθ2cosθ)⇒f(x)=1+tan−1(tanθ)⇒f(x)=1+θ⇒f(x)=1+sin−1(x2)⇒f′(x)=0+1√1−(x2)2⋅12⇒f′(0)=12