**For any two distinct elements x**_{1} and x_{2 }in [0, π/2]

We know that

sin x_{1 }≠ sin x_{2} and cos x_{1} ≠ cos x_{2}

So we get

f(x_{1}) ≠ f(x_{2}) and g(x_{1}) ≠ g(x_{2})

**Here, both f and g are one-one**

We know that

(f + g) (x) = f( x) + g(x) = sin x + cos x

**By substituting the values**

(f + g) (0) = sin 0 + cos 0 = 1

(f + g) (π/2) = sin π/2 + cos π/2 = 1

We know that 0 ≠ π/2

Here, (f + g) (0) = (f + g) (π/2)

**So, f + g is not one-one.**