f:
and g: [ – 1,1]
R defined as f(x) = tan x and g(x) = ![]()
Range of f: let y = f(x)
⇒ y = tan x
⇒ x = tan – 1 y
Since, x ϵ
, y ϵ (– ∞, ∞)
As Range of f ⊂ Domain of g
∴ gof exists.
Similarly, let y = g(x)
⇒ y = ![]()
⇒ x = ![]()
∴ Range of g is [ – 1,1]
As, Range of g ⊂ Domain of f
Hence, fog also exists
Now,
fog(x) = f(g(x)) = f![]()
⇒ fog(x) = tan ![]()
Again,
gof(x) = g(f(x)) = g(tan x)
⇒ gof(x) = ![]()