Concept:Use the identity sin−1x+cos−1x=2π for x>0, then simplify the expression using trigonometric formulas.Explanation:Since x>0 and both sin−1x and cos−1x exist, we have 0<x≤1.So sin−1x+cos−1x=2π.Let A=cos−1x and T=tan−1x.Then sin−1x=2π−A.The left side becomes sin(A+T)−cos(2π−A+T).Now cos(2π−(A−T))=sin(A−T).Thus LHS =sin(A+T)−sin(A−T).Using sinC−sinD=2cos2C+Dsin2C−D, we get:=2cosAsinT.Since cosA=x and sinT=1+x2x,LHS =1+x22x2.Also, sin(cot−12)=51.So 1+x22x2=51.Squaring: 1+x24x4=51.Let y=x2. Then 20y2−y−1=0.(4y−1)(5y+1)=0.Since y>0, y=41.Therefore x2=41 and x=21 because x>0.Answer:x=21, i.e. Option B.