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Evaluate : $\begin{vmatrix} cos 15^{\circ} & sin 15^{\circ} \\ sin 75^{\circ} & cos 75^{\circ} \end{vmatrix} $

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Toolbox:
  • The determinant of a square matrix of order $2 \times 2$ is $|A|=a_{11} \times a_{22} - a_{12} \times a_{21}$
  • $\cos (A+B) =\cos A \cos B- \sin A \sin B$
  • $\cos 90^{\circ}=0$
Let $\Delta=\begin{vmatrix} cos 15^{\circ} & sin 15^{\circ} \\ sin 75^{\circ} & cos 75^{\circ} \end{vmatrix} $
on expanding we get,
$\Delta =\cos 15^{\circ} \cos 75^{\circ}-\sin 15^{\circ}\sin 75^{\circ}$
This is of the form $\cos (A+B)=\cos A \cos B-\sin A\sin B$
Therefore $\cos 15^{\circ} \cos 75^{\circ}-\sin 15^{\circ}\sin 75^{\circ}=\cos (15+75)$
$=\cos 90^{\circ}$
But $\cos 90^{\circ}=0$
Therefore $\Delta=0$
answered Apr 4, 2013 by meena.p
 

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