If C = 2 cos theta,then the value of the dete | vedclass

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(D) Given the determinant $\Delta = \begin{vmatrix} C & 1 & 0 \ 1 & C & 1 \ 6 & 1 & C \end{vmatrix}$.Expanding along the first row:$\Delta = C(C^2 -

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None of these

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(D) Given the determinant $\Delta = \begin{vmatrix} C & 1 & 0 \ 1 & C & 1 \ 6 & 1 & C \end{vmatrix}$.Expanding along the first row:$\Delta = C(C^2 - 1) - 1(C - 6) + 0(1 - 6C)$$\Delta = C^3 - C - C + 6$$\Delta = C^3 - 2C + 6$Substitute $C = 2 \cos 	heta$ into the expression:$\Delta = (2 \cos 	heta)^3 - 2(2 \cos 	heta) + 6$$\Delta = 8 \cos^3 	heta - 4 \cos 	heta + 6$Since $4 \cos^3 	heta - 3 \cos 	heta = \cos 3	heta$,we have $8 \cos^3 	heta = 2 \cos 3	heta + 6 \cos 	heta$.Thus,$\Delta = 2 \cos 3	heta + 6 \cos 	heta - 4 \cos 	heta + 6 = 2 \cos 3	heta + 2 \cos 	heta + 6$.None of the given options match this result.

If $C = 2 \cos \theta$,then the value of the determinant $\Delta = \begin{vmatrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{vmatrix}$ is

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