f(x) = x + 1 and g(x) = sin x

Range of f = R ⊂ Domain of g = R ⇒ gof exists

Range of g= [ – 1,1] ⊂ Domain of f ⇒ fog exists

Now,

fog(x) = f(g(x)) = f(sin x) = sin x + 1

And

gof(x) = g(f(x)) = g(x + 1) = sin(x + 1)

Hence, fog(x) = sin x + 1 and gof(x) = sin(x + 1)