Question

(sina-cosa)^2/sin^4a+2sin^2acos^2a+cos^4a=1-sin2a

Answer 1

(sina-cosa)^2/sin^4a+2sin^2acos^2a+cos^4a=1-sin2a
(sina-cosa)^2 = Sin²α -2SinαCosα + Cos²α = 1 - Sin2α
sin^4a+2sin^2acos^2a+cos^4a =(Sin²2α + Cos²2α)² =1

Answer 2

Let's simplify the expression step by step:

Starting with the left-hand side of the equation:

(sina-cosa)^2 / (sin^4a + 2sin^2acos^2a + cos^4a)

Expanding the numerator:

(sin^2a - 2sina*cosa + cos^2a) / (sin^4a + 2sin^2acos^2a + cos^4a)

Using the trigonometric identity sin^2a + cos^2a = 1:

(1 - 2sina*cosa) / (sin^4a + 2sin^2acos^2a + cos^4a)

Factoring the denominator:

(1 - 2sina*cosa) / [(sin^2a + cos^2a)^2 + 2sin^2acos^2a]

Simplifying further:

(1 - 2sina*cosa) / [1 + 2sin^2acos^2a]

Using the trigonometric identity 2sin^2a*cos^2a = sin^2(2a):

(1 - 2sina*cosa) / [1 + sin^2(2a)]

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

1 - sin(2a)

We have obtained the expression on both sides of the equation in a simplified form. It is not apparent from this point whether the two expressions are equal. To verify their equality, we would need additional information or conditions regarding the variables involved.