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if tan (x+y)/tan (x-y)=2p+q/2p-q, find 1-cos4x/1-cos4y.?

11 months ago

tan(x+y)/tan(x-y)=2p+q/2p-q
so,
if a/b=c/d
then , a+b/ a-b = c+d / c- d
therefore
tan(x+y) + tan (x-y) / tan(x+y) - tan(x-y)
= 2p + q + 2p -q / 2p + q - (2p - q)
Simplifying by breaking "tan" function into sin and cos we get,
sin(x+y) cos(x-y)+sin(x-y) cos (x+y) /sin(x+y)cos(x-y)-sin(x-y)cos (x+y) = 2p/q

Now by applying
sinAcosB+sinBcosA=sin(A+B)
sinAcosB-sinBcosA=sin(A-B)
sin(2x)/sin(2y) = 2p/q
squaring both sides
[Sin(2x)]^2 / [sin(2y)]^2 = 4p^2/q^2
By applying
1-cos4A=sin(2A)^2
1-cos(4x)/1-cos(4y) = 4 p^2 / q^2

10 months ago
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