Verify if (sin A + cos A)^2 + (sin A - cos A)^2 = 2?

We'll have to expand the squares, using the
formulas:

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

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

According to these formulas, we'll
get:

(sin A + cos A)^2 = (sin A)^2 + 2sinA*cosA + (cos A)^2
(1)

(sin A - cos A)^2 = (sin A)^2 - 2sinA*cosA + (cos A)^2
(2)

We'll use the Pythagorean
identity:

 (sin A)^2 + (cos A)^2 =
1

We'll add the developments (1) and
(2);

1 + 2sinA*cosA + 1
- 2sinA*cosA

We'll eliminate like
terms:

(sin A + cos A)^2 + (sin A - cos A)^2
= 1 + 1 = 2