The correct option is C f(x)=sin (cos−1x) and g(x)=cos (sin−1x)
For the two functions to be identical, their range and domain should be same.→f(x) = ln x2 & g(x) = 2 ln x the domain of f is R − (0}, while the domain of g is R+so, f and g are not identical→f(x)=logxe & g(x) = 1logex,the domain of f and g is R+ − (1}and f(x)=logxe = 1logex = g(x)so, f and g are identical→f(x) = sin (cos−1x) & g(x) = cos (sin−1x),the domain of f and g are [−1, 1]and f(x) = sin (cos−1x) =sin (π2−sin−1x) (sin−1x + cos−1x = π2, for x ∈[−1, 1] )=cos(sin−1x)=g(x)so, f and g are identical