Solution
To solve the problem, we will use the trigonometric identities of sin and cos to express tan^2(x y/2) and tan^2(x-y/2) in terms of a and b.
Expressing tan^2(x y/2) in terms of a and b
We know that:
tan(x y/2) = sin(x y)/(cos(x) + cos(y))
Squaring both sides, we get:
tan^2(x y/2) = sin^2(x y)/(cos^2(x) + 2cos(x)cos(y) + cos^2(y))
Using the identity sin^2(x) + cos^2(x) = 1, we can simplify the denominator to:
tan^2(x y/2) = sin^2(x y)/(1 + 2cos(x)cos(y))
We are given that sin(x)sin(y) = a and cos(x)cos(y) = b. Using these values, we can substitute in the expression to get:
tan^2(x y/2) = (a/b)^2 / (1 + 2b)
Expressing tan^2(x-y/2) in terms of a and b
We know that:
tan(x-y/2) = sin(x-y)/(cos(x) + cos(y))
Using the identity sin(x-y) = sin(x)cos(y) - cos(x)sin(y), we can simplify the numerator to:
tan(x-y/2) = (sin(x)cos(y) - cos(x)sin(y))/(cos(x) + cos(y))
Dividing the numerator and denominator by cos(x)cos(y), we get:
tan(x-y/2) = (a/b - 1)/(1 + b)
Squaring both sides, we get:
tan^2(x-y/2) = (a^2 + b^2 - 2ab)/(b^2 + 2b + 1)
Final Solution
We can now substitute the expressions for tan^2(x y/2) and tan^2(x-y/2) in the required expression and simplify to get:
tan^2(x y/2) tan^2(x-y/2) = [(a/b)^2 / (1 + 2b)][(a^2 + b^2 - 2ab)/(b^2 + 2b + 1)]
= a^2 / (b^2 + 2b + 1)
Therefore, tan^2(x y/2) tan^2(x-y/2) = a^2 / (b^2 + 2b + 1)