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Prove that the determinant $$\begin{vmatrix} x & \sin\theta  & \cos\theta  \\ -\sin\theta  & -x & 1 \\ \cos\theta  & 1 & x \end{vmatrix}$$ is independent of $$\theta$$.


Solution

$$\begin{vmatrix} x & \sin\theta  & \cos\theta  \\ -\sin\theta  & -x & 1 \\ \cos\theta  & 1 & x \end{vmatrix}$$

Expanding along the first row, we get
$$\Delta = x(-x^2-1)-\sin\theta(-x\sin \theta-\cos \theta)+\cos \theta(-\sin\theta+x\cos\theta )$$
$$=-x^3-x+x\sin^2\theta+\sin\theta\cos\theta -\sin\theta\cos\theta +x\cos^2\theta$$
$$=-x^3-x +x(\sin^2\theta+\cos^2\theta)$$
$$\Rightarrow \Delta=-x^3$$       ($$\because \sin^2\theta+\cos^2\theta=1$$)
Hence, the given determinant is independent of $$\theta$$


Mathematics
NCERT
Standard XII

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