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)