The value of the determinant left| begin{arra | vedclass

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

Question

(D) Let $D = \left| \begin{array}{ccc} 1 & \cos(\alpha - \beta) & \cos \alpha \ \cos(\alpha - \beta) & 1 & \cos \beta \ \cos \alpha & \cos \beta & 1

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Let $D = \left| \begin{array}{ccc} 1 & \cos(\alpha - \beta) & \cos \alpha \ \cos(\alpha - \beta) & 1 & \cos \beta \ \cos \alpha & \cos \beta & 1 \end{array} ight|$.Expanding along the first row:$D = 1(1 - \cos^2 \beta) - \cos(\alpha - \beta)(\cos(\alpha - \beta) - \cos \alpha \cos \beta) + \cos \alpha(\cos \beta \cos(\alpha - \beta) - \cos \alpha)$.Using the identity $\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$:$D = \sin^2 \beta - \cos(\alpha - \beta)(\sin \alpha \sin \beta) + \cos \alpha(\cos \beta(\cos \alpha \cos \beta + \sin \alpha \sin \beta) - \cos \alpha)$.$D = \sin^2 \beta - \cos(\alpha - \beta)\sin \alpha \sin \beta + \cos \alpha(\cos \alpha \cos^2 \beta + \sin \alpha \sin \beta \cos \beta - \cos \alpha)$.$D = \sin^2 \beta - \cos(\alpha - \beta)\sin \alpha \sin \beta + \cos^2 \alpha \cos^2 \beta + \sin \alpha \sin \beta \cos \alpha \cos \beta - \cos^2 \alpha$.$D = \sin^2 \beta - \cos(\alpha - \beta)\sin \alpha \sin \beta + \cos^2 \alpha(\cos^2 \beta - 1) + \sin \alpha \sin \beta \cos \alpha \cos \beta$.$D = \sin^2 \beta - \sin \alpha \sin \beta(\cos(\alpha - \beta) - \cos \alpha \cos \beta) - \cos^2 \alpha \sin^2 \beta$.$D = \sin^2 \beta - \sin \alpha \sin \beta(\sin \alpha \sin \beta) - \cos^2 \alpha \sin^2 \beta$.$D = \sin^2 \beta - \sin^2 \alpha \sin^2 \beta - \cos^2 \alpha \sin^2 \beta$.$D = \sin^2 \beta - \sin^2 \beta(\sin^2 \alpha + \cos^2 \alpha) = \sin^2 \beta - \sin^2 \beta(1) = 0$.

The value of the determinant $\left| \begin{array}{ccc} 1 & \cos(\alpha - \beta) & \cos \alpha \\ \cos(\alpha - \beta) & 1 & \cos \beta \\ \cos \alpha & \cos \beta & 1 \end{array} \right|$ is

Similar Questions

The value of $\det A$,where $A = \begin{bmatrix} 1 & \cos \theta & 0 \\ -\cos \theta & 1 & \cos \theta \\ -1 & -\cos \theta & 1 \end{bmatrix}$,lies

The value of $\left| \begin{array}{ccc} a & a + b & a + 2b \\ a + 2b & a & a + b \\ a + b & a + 2b & a \end{array} \right|$ is equal to

If $5$ is one root of the equation $\left| \begin{array}{ccc} x & 3 & 7 \\ 2 & x & -2 \\ 7 & 8 & x \end{array} \right| = 0$,then the other two roots of the equation are:

If one of the roots of $\left|\begin{array}{lll}3 & 5 & x \\ 7 & x & 7 \\ x & 5 & 3\end{array}\right|=0$ is $-10$,then the other roots are

The value of $\begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module