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) =