Verify/prove that:

sinAcosA/1-cosA = sin^2/tanA-sinA

tanA-sinA = sin(1/cos - 1)

see where that takes you.

To verify/prove the given equation:

sinAcosA / (1 - cosA) = sin^2A / (tanA - sinA)

Let's start by simplifying the left side of the equation:

sinAcosA / (1 - cosA)

Using the identity sin2A = 2sinAcosA, we can rewrite sinAcosA as (1/2) * sin2A:

(1/2) * sin2A / (1 - cosA)

Next, let's simplify the right side of the equation:

sin^2A / (tanA - sinA)

Using the identity tanA = sinA / cosA, we can rewrite the denominator as cosA * tanA - sinA:

sin^2A / (cosA * tanA - sinA)

Using the identity sin^2A = 1 - cos^2A, we can rewrite sin^2A as 1 - cos^2A:

(1 - cos^2A) / (cosA * tanA - sinA)

Now, we can see that the left side and right side of the equation have the same form:

(1/2) * sin2A / (1 - cosA) = (1 - cos^2A) / (cosA * tanA - sinA)

To prove that they are equal, it is enough to show that their numerators and denominators are equal:

Numerator:
(1/2) * sin2A = 1 - cos^2A

Since sin2A = 2sinAcosA:
1 = 2sinAcosA + cos^2A
1 = cos^2A + sinAcosA + sinAcosA
1 = cos^2A + 2sinAcosA
1 = cos^2A + sin2A

Using the double angle identity for sine (sin2A = 2sinAcosA):
1 = cos^2A + sin2A

This shows that the numerators are equal.

Denominator:
(1 - cosA) = (cosA * tanA - sinA)

Expanding the right side using the identity tanA = sinA / cosA:
1 - cosA = sinA - sinA = 0

Therefore, the denominators are equal.

Since the numerators and denominators of the left side and right side of the equation are equal, we have verified/proven that:

sinAcosA / (1 - cosA) = sin^2A / (tanA - sinA)