Let `f(x)=sin^2(x/2) +cos^2(x/2)` and `g(x)=sec^2 x- tan^2 x`. The two functions are equal over the set (A) `phi` (B) `R` (C) `R-{0}` (D) 1

Let f(x) = sin^2(x//2) + cos^2(x//2) and g(x) = sec^2x- tan^2x . The two function are equal over the set :

Let f(x)=sin^(2)((x)/(2))+cos^(2)((x)/(2)) and g(x)=sec^(2)x-tan^(2)x. The two functions are equal over the set (A) phi( B) R( ) R-{0} (D) 1

38. Let/(x)=sin'. + cos" ec x -tan x.The two functions g ( x ) = sec x tan are equal over the set 1) ? 2) R 3pR ?tx|x=(2n + 1 )?.neZ} 4)R-{? 2

f(x)=(cos^(2)x)/(1+cos x+cos^(2)x) and g(x)=k tan x+(1-k)sin x-x, where k in R,g'(x)=

Let f(x)=x^2 and g(x)=2^x . Then the solution set of the equation fog(x)=gof(x) is R (b) {0} (c) {0, 2} (d) none of these

For all x in R, let f(x)=sin^(4)x+cos^(4)x and g(x)=sin^(6)x+cos^(6)x, then (3f(x)-2g(x)) is equal to

Let (b cos x)/(2 cos 2x-1)=(b + sin x)/((cos^(2) x-3 sin^(2) x) tan x), b in R . Equation has solutions if

Let f(x) = x(2-x), 0 le x le 2 . If the definition of f(x) is extended over the set R-[0,2] by f (x+1)= f(x) , then f is a

Let f (x) = int x ^(2) cos ^(2)x (2x + 6 tan x - 2x tan ^(2) x ) dx and f (x) passes through the point (pi, 0) If f : R-(2n +1) pi/2 to R then f (x) be a :

Let f(x)=sin^(-1)((2x)/(1+x^(2)))AA x in R. The function f(x) is continuous everywhere but not differentiable at x is/ are