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

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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**

Q: Prove : (sin A + cos A)² + (sin A - cos A)² =2

A: L:H:S ≡ (sin A + cos A)² + (sin A - cos A)²

= sin² A + cos²A +2sinAcos A + sin²A + cos²A - 2sinAcosA

{ Using (A+B)² = A²+B²+2AB , (A-B)= A²+B²-2AB }

= 1 + 2sinAcosA + 1 - 2sinAcosA

{ Using sin²A + cos² A = 1 }

= 2

∴ L:H:S≡R:H:S

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