Expanding the left side of the equation, we have:
(1 + sinα – cosα)² + (1 – sinα + cosα)²
= (1 + sinα – cosα)(1 + sinα – cosα) + (1 – sinα + cosα)(1 – sinα + cosα)
= 1 + 2sinα - 2cosα + sin²α - 2sinαcosα + cos²α + 1 - 2sinα + 2cosα + sin²α - 2sinαcosα + cos²α
= 2 + 2sin²α - 4sinαcosα + 2cos²α
Using the trigonometric identity sin²α + cos²α = 1, we can rewrite the expression as:
2 + 2(1 - cos²α) - 4sinαcosα + 2cos²α
= 2 + 2 - 2cos²α - 4sinαcosα + 2cos²α
= 4 - 2cos²α - 4sinαcosα
Now, let's simplify the right side of the equation:
4(1 – sinα . cosα)
= 4 - 4sinαcosα
From here, we can see that the left side and the right side of the equation are equal, as both simplify to 4 - 2cos²α - 4sinαcosα.
Therefore, the given equation is true for all values of α.
(o) (1 + sinα – cosα)2 + (1 – sinα + cosα)2 = 4(1 – sinα . cosα)
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