(ii) sin2α+sin2β+cos2(α+β)+2sinαsinβcos(α+β)=1\sin^2\alpha + \sin^2\beta + \cos^2(\alpha + \beta) + 2\sin\alpha\sin\beta\cos(\alpha + \beta) = 1sin2α+sin2β+cos2(α+β)+2sinαsinβcos(α+β)=1