Question

$\begin{vmatrix} {Sin \alpha}&{ \cos\alpha} &Sin ({\alpha+ \delta })\\\ {Sin \beta}&{ Cos \beta}& Sin ({\beta+\delta}) \\\ {Sin \gamma}&{ Cos \gamma}&Sin ({\gamma+\delta})\\\ \end{vmatrix}$ is equal to

• A. $0$
• B. $1$
• C. $1+ Sin \alpha\, Sin \beta\, Sin \gamma$
• D. $1-(Sin \alpha - Sin \beta) (Sin \beta-Sin \gamma)(Sin \gamma - Sin \alpha)$

Answer 1

Answer: (A)

$0$

Answer 2

Given, $\begin{vmatrix}\sin \alpha & \cos \alpha & \sin (\alpha+\delta) \\\ \sin \beta & \cos \beta & \sin (\beta+\delta) \\\ \sin \gamma & \cos \gamma & \sin (\gamma+\delta)\end{vmatrix}$ $= \begin{vmatrix}\sin \alpha & \cos \alpha & \sin \alpha \cdot \cos \delta+\cos \alpha \cdot \sin \delta \\\ \sin \beta & \cos \beta & \sin \beta \cdot \cos \delta+\cos \beta \cdot \sin \delta \\\ \sin \gamma & \cos \gamma & \sin \gamma \cdot \cos \delta+\cos \gamma \cdot \sin \delta\end{vmatrix}$ $= \begin{vmatrix}\sin \alpha & \cos \alpha & \sin \alpha \cdot \cos \delta \\\ \sin \beta & \cos \beta & \sin \beta \cdot \cos \delta \\\ \sin \gamma & \cos \gamma & \sin \gamma \cdot \cos \delta\end{vmatrix}$ $+ \begin{vmatrix}\sin \alpha & \cos \alpha & \cos \alpha \cdot \sin \delta \\\ \sin \beta & \cos \beta & \cos \beta \cdot \sin \delta \\\ \sin \gamma & \cos \gamma & \cos \gamma \cdot \sin \delta\end{vmatrix}$ $=\cos \delta \begin{vmatrix}\sin \alpha & \cos \alpha & \sin \alpha \\\ \sin \beta & \cos \beta & \sin \beta \\\ \sin \gamma & \cos \gamma & \sin \gamma\end{vmatrix}$ $+\sin \delta \begin{vmatrix}\sin \alpha & \cos \alpha & \cos \alpha \\\ \sin \beta & \cos \beta & \cos \beta \\\ \sin \gamma & \cos \gamma & \cos \gamma\end{vmatrix}$ $=\cos \delta \times 0+\sin \delta \times 0$ $\left(\because C_{1}\right.$ and $C_{2}$ are identical $C_{2}$ and $C_{3}$ identical.) $=0$