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  4. The determinant |x sin θ cos θ - sin θ -x 1 cos θ 1 x| is

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Q. The determinant $\begin{vmatrix}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{vmatrix}$ is

Determinants

Solution:

Let $\Delta=\begin{vmatrix}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{vmatrix}$
By expanding along first row, we get
$\Delta=x\begin{vmatrix}-x & 1 \\ 1 & x\end{vmatrix}-\sin \theta\begin{vmatrix}-\sin \theta & 1 \\ \cos \theta & x\end{vmatrix}+\cos \theta\begin{vmatrix}-\sin \theta & -x \\ \cos \theta & 1\end{vmatrix}$
$= x\left(-x^2-1\right)-\sin \theta(-x \sin \theta-\cos \theta) +\cos \theta(-\sin \theta+x \cos \theta) $
$= -x^3-x+x \sin ^2 \theta+\sin \theta \cos \theta-\sin \theta \cos \theta+x \cos ^2 \theta $
$=-x^3-x+x\left(\sin ^2 \theta+\cos ^2 \theta\right)= -x^3-x+x $
$ \left(\because \sin ^2 \theta+\cos \theta^2=1\right)$
$= -x^3 $
Hence, $\Delta$ is independent of $\theta$.