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

In order that the matrix $\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 & \lambda & 5 \end{bmatrix}$ be non-singular,$\lambda$ should not be equal to:

If $\alpha, \beta, \gamma$ $(\alpha < \beta < \gamma)$ are the values of $x$ such that $\begin{vmatrix} x-2 & 0 & 1 \\ 1 & x+3 & 2 \\ 2 & 0 & 2x-1 \end{vmatrix} = 0$ is a singular matrix,then $2\alpha + 3\beta + 4\gamma = $

If $f(x) = \begin{vmatrix} 1 & x & x+1 \\ 2x & x(x-1) & (x+1)x \\ 3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1) \end{vmatrix}$, then $f(100)$ is equal to:

If $\omega$ is the cube root of unity,then $\left| \begin{array}{ccc} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{array} \right| = $

If ${a^{ - 1}} + {b^{ - 1}} + {c^{ - 1}} = 0$ such that $\left| {\begin{array}{*{20}{c}}{1 + a}&1&1\\1&{1 + b}&1\\1&1&{1 + c}\end{array}} \right| = \lambda $,then the value of $\lambda $ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module