To prove what is required in this question first open the brackets and find the square of ( sin x + cos x )

( sin x + cos x ) ^2 = (sin x )^2 + ( cos x)^2 + 2 sin x * cos x

Now we have to eliminate ( sin x )^2 and (cos x )^2 .

To do this we use a relation that we already know which is: (sin x )^2 + ( cos x)^2 =1 .

Substituting this in (sin x )^2 + ( cos x)^2 + 2 sin x * cos x we arrive at 1 + 2 sin x * cos x

**Therefore ( sin x + cos x ) ^2 = 1 + 2 sin x * cos x**

To prove that (sinx+cosx)^2 = 1+2sinxcosx.

We kanow that (sinx)^2+ (cosx)^2 = 1 is trigonometric identy.

Therefore

LHS:

(sinx+cosx)^2 = (sinx)^2 +2sinx*cosx + (cosx)^2

(sinx+cosx)^2 = (sinx)^2 + (cosx)^2 +2sinx*cosx

(sinx+cosx)^2 = 1 +2sinx*cosx , as (sinx)^2+(cosx)^2 = 1.

Therefore (sinx+cosx)^2 = 1 +2sinx*cosx = RHS.

To prove the equality means to obtain like expressions both sides.

We'll start by expanding the square from the left side. We'll use the formula of expanding the square:

(a+b)^2 = a^2 + 2ab + b^2

We'll substitute a and b by sin x and cos x and we'll get:

( sinx + cosx )^2 = (sin x)^2 + 2sin x*cos x + (cos x)^2

But the sum (sin x)^2 + (cos x)^2 = 1, from the fundamental formula of trigonometry.

We'll substitute the sum by it's value. We'll re-write the expanding of the square:

**( sinx + cosx )^2 = 1 + 2sin x*cos x q.e.d**