if x=cotA+cosA and cotA-cosA then (x2-y2)2=4xy
Yash0804:
Is that y = cotA - cosA
ss
and is that (x^2 - y^2)^2
ss
it answer may be 16xy
But question is to find the value of (x^2-y^2)^2=4xy then why the answer will be in xy it would be in terms of cosA and cotA
I think you are asking for proof
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(x^2-y^2)^2 = 4xy
[(x+y)(x-y)]^2 = 4xy
(4cotAcosA)^2 = 4[cot^2(A) - cos^2(A)]
16cot^2(A)cos^2(A) = 4[cot^2(A) - cos^2(A)]
Solving the right side
= 4[ { cos^2(A)/sin^2(A)} - cos^2(A)]
Taking cos^2(A) common
4cos^2(A) [ { 1/sin^2(A) } - 1 ]
4cos^2(A)/sin^2(A) [1 - sin^2(A)]
4cot^2(A)cos^2(A)
I think instead of 4xy it should be 16xy in order to proof LHS = RHS
(x^2-y^2)^2 = 4xy
[(x+y)(x-y)]^2 = 4xy
(4cotAcosA)^2 = 4[cot^2(A) - cos^2(A)]
16cot^2(A)cos^2(A) = 4[cot^2(A) - cos^2(A)]
Solving the right side
= 4[ { cos^2(A)/sin^2(A)} - cos^2(A)]
Taking cos^2(A) common
4cos^2(A) [ { 1/sin^2(A) } - 1 ]
4cos^2(A)/sin^2(A) [1 - sin^2(A)]
4cot^2(A)cos^2(A)
I think instead of 4xy it should be 16xy in order to proof LHS = RHS
ss
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