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(D) Let the determinant be $\Delta = \left| \begin{matrix} \sin \alpha & \cos \alpha & \sin(\alpha + \gamma) \ \sin \beta & \cos \beta & \sin(\beta +

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(D) Let the determinant be $\Delta = \left| \begin{matrix} \sin \alpha & \cos \alpha & \sin(\alpha + \gamma) \ \sin \beta & \cos \beta & \sin(\beta + \gamma) \ \sin \delta & \cos \delta & \sin(\delta + \gamma) \end{matrix} ight|$.Using the trigonometric identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we can expand the third column:$\sin(\alpha + \gamma) = \sin \alpha \cos \gamma + \cos \alpha \sin \gamma$$\sin(\beta + \gamma) = \sin \beta \cos \gamma + \cos \beta \sin \gamma$$\sin(\delta + \gamma) = \sin \delta \cos \gamma + \cos \delta \sin \gamma$Now,the determinant becomes:$\Delta = \left| \begin{matrix} \sin \alpha & \cos \alpha & \sin \alpha \cos \gamma + \cos \alpha \sin \gamma \ \sin \beta & \cos \beta & \sin \beta \cos \gamma + \cos \beta \sin \gamma \ \sin \delta & \cos \delta & \sin \delta \cos \gamma + \cos \delta \sin \gamma \end{matrix} ight|$.Apply the column operation $C_3 ightarrow C_3 - (\cos \gamma) C_1 - (\sin \gamma) C_2$:Each element in the third column becomes $(\sin 	heta \cos \gamma + \cos 	heta \sin \gamma) - (\cos \gamma)(\sin 	heta) - (\sin \gamma)(\cos 	heta) = 0$.Since all elements of the third column are $0$,the value of the determinant is $\Delta = 0$.

The value of $\left| \begin{matrix} \sin \alpha & \cos \alpha & \sin(\alpha + \gamma) \\ \sin \beta & \cos \beta & \sin(\beta + \gamma) \\ \sin \delta & \cos \delta & \sin(\delta + \gamma) \end{matrix} \right|$ is

Similar Questions

Let $A$ be a $3 \times 3$ matrix with $\operatorname{det}(A) = 4$. Let $R_{i}$ denote the $i^{\text{th}}$ row of $A$. If a matrix $B$ is obtained by performing the operation $R_{2} \rightarrow 2R_{2} + 5R_{3}$ on $2A$,then $\operatorname{det}(B)$ is equal to:

By using properties of determinants,show that:
$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$

By using properties of determinants,show that:
$\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$

Let $P = [a_{ij}]$ be a $3 \times 3$ matrix and let $Q = [b_{ij}]$,where $b_{ij} = 2^{i+j} a_{ij}$ for $1 \leq i, j \leq 3$. If the determinant of $P$ is $2$,then the determinant of the matrix $Q$ is:

The product of a matrix and its transpose is an identity matrix. The value of the determinant of this matrix is:

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