Show that: `sin^2 alpha + sin^2 beta + 2sinalpha sinbeta cos(alpha+beta)=sin^2 (alpha+beta)`

Explanation

Let `alpha+beta=theta`
Taking cos on both sides,
`cos(alpha+beta)=cos theta`
`rArr cosalpha cosbeta= sinalpha sin beta + cos theta`
Squarring both sides, we obtain
`cos^(2)alpha cos^(2)beta= sin^(2)alpha sin^(2)beta+2 sin alpha sinbeta costheta + cos^(2)theta`
`rArr (1-sin^(2)alpha)(1-sin^(2)beta)=sin^(2)alphasin^(2)beta=sin^(2)alpha sin^(2)beta+2sinalphasinbetacostheta+1-sin^(2)theta`
`rArr sin^(2)alpha+sin^(2)beta+2sinalphasinbetacos(alpha+beta) =sin^(2)(alpha+beta)`