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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 -

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(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

Similar Questions

If $a \neq b \neq c$,the value of $x$ which satisfies the equation $\left| \begin{array}{ccc} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{array} \right| = 0$ is

The area of $\triangle PQR$ with the vertices $P(k, 1)$,$Q(2, 4)$,and $R(1, 1)$ is $3$ sq. units. Then,$k = $ . . . . . . .

The set of all values of $\lambda$ for which the system of linear equations $2x_1 - 2x_2 + x_3 = \lambda x_1$,$2x_1 - 3x_2 + 2x_3 = \lambda x_2$,and $-x_1 + 2x_2 = \lambda x_3$ has a non-trivial solution:

$\left|\begin{array}{ccc} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{array}\right|=$

If $a, b, c$ are in $A.P.$,then the value of $\left| \begin{array}{ccc} x+2 & x+3 & x+a \\ x+4 & x+5 & x+b \\ x+6 & x+7 & x+c \end{array} \right|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module