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STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 3 and 4 . determinant and metrices1 Mark
MCQ
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
  • A
    ${\alpha ^2} + {\beta ^2}$
  • B
    ${\alpha ^2} - {\beta ^2}$
  • C
    $1$
  • ✓
    $0$
✓

Answer

Correct option: D.
$0$
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 $

= $1 - {\cos ^2}\beta - {\sin ^2}\beta  = 0.$

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