If `cos (theta - alpha) = a and sin(theta - beta) = b (0 lt theta - alpha, theta - beta lt pi//2)`, then prove that `cos^(2) (alpha - beta) + 2ab sin (alpha - beta) = a^(2) + b^(2)`

Explanation

`cos (theta - alpha) = a and sin (theta - beta) = b`
`theta = alpha + cos^(-1) a = beta + sin^(-1) b " " ( :' 0 lt theta - alpha, theta - beta lt pi//2)`
`:. alpha - beta = sin^(-1) b - cos^(-1) a`
`rArr sin (alpha - beta) = sin (sin^(-1) b - cos^(-1) a)`
`rArr sin (alpha - beta) = sin (sin^(-1) b - cos^(-1) a)`
`rArr sin(alpha - beta) = ab - sqrt(1 -b^(2)) sqrt(1 -a^(2))`
`rArr sin (alpha - beta) - ab = - sqrt(1 - b^(2)) sqrt(1- a^(2))`
`rArr sin^(2) (alpha - beta) - 2ab sin (alpha - beta) = 1 - (a^(2) + b^(2))`
`rArr cos^(2) (alpha - beta) + 2 ab sin (alpha - beta) = a^(2) + b^(2)`