- Home
- Standard 12
- Mathematics
The values of the determinant $\left| {\,\begin{array}{*{20}{c}}1&{\cos (\alpha - \beta )}&{\cos \alpha }\\{\cos (\alpha - \beta )}&1&{\cos \beta }\\{\cos \alpha }&{\cos \beta }&1\end{array}\,} \right|$ is
${\alpha ^2} + {\beta ^2}$
${\alpha ^2} - {\beta ^2}$
$1$
$0$
Solution
(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 $
= $1 – {\cos ^2}\beta – {\sin ^2}\beta = 0.$
