Trigonometric identities:
# tan (A+-B) = (tan A +- tan B)/(1 ""_+^(-) tan Atan B) #
# sin^2 A + cos ^A = 1 #
# tan^2 A + 1 = sec^2 A = 1/cos^2 A #
# sin 2A = cos 2(A-pi/2) = -cos 2A#
Solution:
# (a+b)tan (X-Y) = (a-b)tan (X+Y) #
# (a+b)((tan X - tan Y)/(1 + tan X tan Y)) = (a-b)((tan X + tan Y)/(1 - tan X tan Y)) #
# (a+b)(tan X - tan Y)(1 - tan X tan Y) = (a-b)(tan X + tan Y)(1 + tan X tan Y) #
# a tan Y(tan^2 X + 1) = b tan X(tan^2 Y + 1) #
# a sin 2Y = b sin 2X # ...(Eq. 1)
Now it is possible to determine that:
# a cos 2Y = b cos 2X # ...(Eq. 2)
Summing the squares of equations 1 and 2:
# a^2 (sin^2 2Y + cos^2 2Y) = b^2 (sin^2 2X + cos^2 2X) #
# a^2 = b^2 #
From question:
# a cos 2Y + b cos 2X = c #
sum equation 1 to obtain:
# 2a cos 2Y = c #
Multiply by # c # and add # a^2-b^2 = 0 # to obtain:
# a^2 - b^2 + c^2 = 2ac cos 2Y #
