(o(1 + sinα – cosα)^2 + (1 – sinα + cosα)^2 = 4(1 – sinα . cosα)

Answer 1

Let's simplify the left-hand side of the equation first:

(1 + sinα – cosα)^2 + (1 – sinα + cosα)^2
= (1 + sinα – cosα)(1 + sinα – cosα) + (1 – sinα + cosα)(1 – sinα + cosα)
= (1 + sinα – cosα)(1 + sinα – cosα) + (1 – sinα + cosα)(1 – sinα + cosα)
= (1 + sinα)(1 + sinα) + (1 – sinα)(1 – sinα) – (cosα)(cosα) + (cosα)(cosα)
= 1 + 2sinα + sinα^2 + 1 - 2sinα + sinα^2 - cosα^2 + cosα^2
= 2 + 2sinα^2

Now let's simplify the right-hand side of the equation:

4(1 – sinα . cosα)
= 4(1 – sinα cosα)
= 4 - 4sinα cosα

Now we can equate both sides of the equation:

2 + 2sinα^2 = 4 - 4sinα cosα

Subtract 2 from both sides:

2sinα^2 = 2 - 4sinα cosα

Divide both sides by 2:

sinα^2 = 1 - 2sinα cosα

Rearrange the equation:

sinα^2 + 2sinα cosα - 1 = 0

We can use the quadratic formula to solve for sinα:

sinα = (-2cosα ± √(4cosα^2 - 4(-1)(1))) / (2)
sinα = (-2cosα ± √(4cosα^2 + 4)) / 2
sinα = -cosα ± √(cosα^2 + 1)

This means that the original equation holds true for any value of α, as long as sinα = -cosα ± √(cosα^2 + 1).