The correct option is B ±√53
Given that tan{cos−1(x)}=sin(cot−112)
Let cot−112=ϕ⇒12=cotϕ
⇒sinϕ=1√1+cot2 ϕ=2√5
Let cos−1x=θ⇒sec θ1x⇒tan θ=√sec2 θ−1
⇒tan θ=√1x2−1⇒tan θ=√1−x2x
So,tan{cos−1(x)}=sin(cot−112)
⇒tan(tan−1√1−x2x)=sin(sin−12√5)⇒√1−x2x=2√5⇒√(1−x2)5=2x
Squaring both sides, we get x=±√53.