If `cos(theta-alpha) = a` and `cos(theta-beta) = b` then the value of `sin^2(alpha-beta) + 2ab cos(alpha-beta)`

If cos (theta - alpha) = a, cos (theta - beta) = b , then the value of sin^(2) (alpha - beta) + 2ab cos (alpha - beta) is

s(theta-alpha)=a and cos(theta-beta)=b, then sin^(2)(alpha-beta)+2ab cos(alpha-beta)=

If sin (theta + alpha) = a and sin (theta + beta) = b,"then" cos 2 (alpha - beta) - 4ab cos (alpha - beta) =

If cos (theta -alpha) =a, sin (theta - beta) =b, then cos ^(2) (alpha - beta) + 2 ab sin (alpha - beta) is equal to

If cos (theta -alpha) =a, sin (theta - beta) =b, then cos ^(2) (alpha - beta) + 2 ab sin (alpha - beta) is equal to

It is given that cos(theta-alpha)=a, cos(theta-beta)=b What is sin^(2)(alpha-beta)+2abcos(alpha-beta) equal to ?

If cos(theta - alpha)=a, cos(theta-beta)=b , then sin^(2)(alpha -beta) + 2abcos(alpha-beta)=0

If cos(theta-alpha)=a , sin(theta-beta)=b then cos^2(alpha-beta)+2 ab sin(alpha-beta) is equal to