Which pair(s) of function(s) is/are equal? `f(x) = cos(2tan^-1 x) ;g(x)=(1-x^2)/(1+x^2)`

Which of the following pairs of functions is/are identical? (a) f(x)="tan"(tan^(-1)x)a n dg(x)="cot"(cot^(-1)x) (b) f(x)=sgn(x)a n dg(x)=sgn(sgn(x)) (c) f(x)=cot^2xdotcos^2xa n dg(x)=cot^2x-cos^2x (d) f(x)=e^(lnsec^(-1)x)a n dg(x)=sec^(-1)x

Find the values of x for which the following pair of functions are identical. (i) f(x)=tan^(-1)x+cot^(-1)x " and " g(x)=sin^(-1)x +cos^(-1)x (ii) f(x)=cos(cos^(-1)x) " and " g(x)=cos^(-1)(cosx)

Examine whether following pair of functions are identical or not? (i) f(x)=(x^(2)-1)/(x-1) and g(x)=x+1 (ii) f(x)=sin^(2)x+cos^(2)x and g(x)=sec^(2)x-tan^(2)x

Find the value of x for which function are identical. f(x)=tan^(-1)x+tan^(-1)( 1/x ) and g(x)=sin^(-1)x+cos^(-1)x

If f(x)=tan^(-1)x+cos^(-1)((1-x^(2))/(1+x^(2))) , then

Check whether following pairs of function are identical or not? f(x)=sqrt((1+cos2x)/2) and g(x)=cosx

Let f(x) be the function given by f(x)=arc cos ((1-x^(2))/(1+x^(2))).Then

Find the value of x for which function are identical. f(x)=cosx and g(x)=1/(sqrt(1+tan^2x))

Column I: Function, Column II: Value of x for which both the functions in any option of column I are identical f(x)=tan^(-1)((2x)/(1-x^2)),g(x)=2tan^(-1)x , p. x in {-1,1} f(x)=sin^(-1)(sinx)a n dg(x)="sin"(sin^(-1)x) , q. x in [-1,1] f(x)=(log)_(x2)25a n dg(x)=(log)_x5 , r. x in (-1,1) f(x)=sec^(-1)x+cos e c^(-1)x ,g(x)=sin^(-1)x+cos^(-1)x , s. x in (0,1)

If x in R , then f(x)=cos^(-1)((1-x^(2))/(1+x^(2))) is equal to