# If Cos4A/ cos2B + sin4A/cos2B = 1 Then prove that: Cos4B/cos2A + sin4B/sin2A = 1 A and B being acute angles of right angled triangle.

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Since the fractions`(cos(4A))/(cos(2B))` and `(sin(4A))/(cos(2B))` have the same denominators, hence, you may write the expression such that:

`(cos(4A) + sin(4A))/(cos(2B)) = 1`

`cos 2*2A + sin 2*2A = cos 2B`

`2cos^2 2A -1 + 2sin2A*cos2A = cos 2B`

`2(2cos^2A - 1)^2 - 1 + 4sinA*cosA*(2cos^2A - 1) = cos2B`

You need to factor out `2(2cos^2A - 1)` such that:

`2(2cos^2A - 1)(2cos^2A - 1 + 2sinA*cosA) = cos2B + 1`

`2(2cos^2A - 1)(2cos^2A - 1 + 2sinA*cosA) = 2cos^2B - 1 + 1`

`2(2cos^2A - 1)(2cos^2A - 1 + 2sinA*cosA) = 2cos^2B`

`(2cos^2A - 1)(2cos^2A - 1 + 2sinA*cosA) = cos^2B`

Since A and B are the acute angles of a right angle triangle yields:

`cos B = sin A => cos^2B= sin^2A`

`(1 - 2sin^2A)(1 - 2sin^2A + 2sinA*cosA) != sin^2A`

**Notice that under the given conditions, the expression `(1 - 2sin^2A)(1 - 2sin^2A + 2sinA*cosA) != sin^2A` proves that `(cos(4A) + sin(4A))/(cos(2B)) = 1` is invalid.**