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

The area of a triangle is $4$ sq. units,and its vertices are $(-2, 0)$,$(0, 4)$,and $(0, k)$. Find the value of $k$.

If $1$,$\log_{10}(4^{x}-2)$ and $\log_{10}(4^{x}+\frac{18}{5})$ are in arithmetic progression for a real number $x$,then the value of the determinant $\left|\begin{array}{ccc} 2(x-\frac{1}{2}) & x-1 & x^{2} \\ 1 & 0 & x \\ x & 1 & 0 \end{array}\right|$ is equal to ...... .

If $2x + 3y - 5z = 7$,$x + y + z = 6$,and $3x - 4y + 2z = 1$,then $x =$

Evaluate the determinant: $\left|\begin{array}{rrr}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|$

The value of $\left| \begin{array}{ccc} 1 & 1 & 1 \\ bc & ca & ab \\ b+c & c+a & a+b \end{array} \right|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module