d
(d) $1\,(1 - {\cos ^2}\beta ) - \cos (\alpha - \beta )$$[\cos (\alpha - \beta ) - \cos \alpha \cos \beta ]$
$ + \cos \alpha [\cos \beta \cos (\alpha - \beta ) - \cos \alpha ]$
$ = 1 - {\cos ^2}\beta - {\cos ^2}\alpha - {\cos ^2}(\alpha - \beta )$$ + 2\cos \alpha \cos \beta \cos (\alpha - \beta )$
= $1 - {\cos ^2}\beta - {\cos ^2}\alpha + \cos (\alpha - \beta )$
$(2\cos \alpha \cos \beta - \cos (\alpha - \beta ))$
= $1 - {\cos ^2}\beta - {\cos ^2}\alpha + \cos (\alpha - \beta )$
$[\cos (\alpha + \beta ) + \cos (\alpha - \beta ) - \cos (\alpha - \beta )]$
= $1 - {\cos ^2}\beta - {\cos ^2}\alpha + \cos (\alpha - \beta )\cos (\alpha + \beta )$
= $1 - {\cos ^2}\beta - {\cos ^2}\alpha + {\cos ^2}\alpha {\cos ^2}\beta - {\sin ^2}\alpha {\sin ^2}\beta $
= $1 - {\cos ^2}\beta - {\cos ^2}\alpha (1 - {\cos ^2}\beta ) - {\sin ^2}\alpha {\sin ^2}\beta $
= $1 - {\cos ^2}\beta - {\cos ^2}\alpha {\sin ^2}\beta - {\sin ^2}\alpha {\sin ^2}\beta $
= $1 - {\cos ^2}\beta - {\sin ^2}\beta = 0.$