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         (i)      \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta &\cos \theta \end{vmatrix}

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The given two by two determinant is calculated as follows

\dpi{100} \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta &\cos \theta \end{vmatrix} = cos \theta (\cos \theta) - (-\sin \theta)\sin \theta = \cos^2\theta + \sin ^2 \theta = 1

Posted by

Divya Prakash Singh

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