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

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

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

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

If sin(theta+alpha)=a and sin(theta+beta)=b then cos2(alpha-beta)-4abcos(alpha-beta) is equal to

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

If sin(theta+alpha)=a and sin(theta+beta)=b, prove that cos2(alpha-beta)-4ab cos(alpha-beta)=1-2a^(2)-2b^(2)

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