prove that (cos2A+cos2B)/(cos2B-cos2A)=cot(A+B)+cot(A-B)
Question
prove that (cos2A+cos2B)/(cos2B-cos2A)=cot(A+B)+cot(A-B)
Solution
Sure, let's prove the given trigonometric identity step by step.
Given: (cos2A + cos2B) / (cos2B - cos2A) = cot(A+B) + cot(A-B)
We know that cos2A = 2cos^2A - 1 and cos2B = 2cos^2B - 1.
Substitute these values into the equation:
= [(2cos^2A - 1) + (2cos^2B - 1)] / [(2cos^2B - 1) - (2cos^2A - 1)] = [2cos^2A + 2cos^2B - 2] / [2cos^2B - 2cos^2A] = 2[cos^2A + cos^2B - 1] / 2[cos^2B - cos^2A] = [cos^2A + cos^2B - 1] / [cos^2B - cos^2A]
We also know that cos^2A + cos^2B = 1 - 2sinAsinB (from the identity cos(A+B)cos(A-B) = cos^2A + cos^2B)
Substitute this into the equation:
= [1 - 2sinAsinB - 1] / [1 - 2sinAsinB - cos^2A] = -2sinAsinB / [1 - 2sinAsinB - cos^2A]
We also know that cos^2A = 1 - sin^2A (from the Pythagorean identity)
Substitute this into the equation:
= -2sinAsinB / [1 - 2sinAsinB - (1 - sin^2A)] = -2sinAsinB / [-2sinAsinB + sin^2A] = 2sinAsinB / [2sinAsinB - sin^2A]
We also know that cot(A+B) = cos(A+B) / sin(A+B) and cot(A-B) = cos(A-B) / sin(A-B)
Add these two equations:
cot(A+B) + cot(A-B) = [cos(A+B) + cos(A-B)] / [sin(A+B) + sin(A-B)] = [2cosAcosB - 2sinAsinB + 2cosAcosB + 2sinAsinB] / [2cosAsinB + 2cosBsinA] = 2cosAcosB / 2cosAsinB = cosAcosB / sinAsinB
This is equal to the left-hand side of the equation, so the given identity is proven.
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