Q120 / 180 · KCET 2011 · Mathematics
$\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin (\alpha+\delta) \\ \sin \beta & \cos \beta & \sin (\beta+\delta) \\ \sin \gamma & \cos \gamma & \sin (\gamma+\delta)\end{array}\right|$ is equal to
- Correct answerA0
- B1
- C$1+\sin \alpha \sin \beta \sin \gamma$
- D$1-(\sin \alpha-\sin \beta)(\sin \beta-\sin \gamma)$ $(\sin \gamma-\sin \alpha)$
Solution
Given, $\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin (\alpha+\delta) \\ \sin \beta & \cos \beta & \sin (\beta+\delta) \\ \sin \gamma & \cos \gamma & \sin (\gamma+\delta)\end{array}\right|$
$=\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin \alpha \cdot \cos \delta+\cos \alpha \cdot \sin \delta \\ \sin \beta & \cos \beta & \sin \beta \cdot \cos \delta+\cos \beta \cdot \sin \delta \\ \sin \gamma & \cos \gamma & \sin \gamma \cdot \cos \delta+\cos \gamma \cdot \sin \delta\end{array}\right|$
$=\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin \alpha \cdot \cos \delta \\ \sin \beta & \cos \beta & \sin \beta \cdot \cos \delta \\ \sin \gamma & \cos \gamma & \sin \gamma \cdot \cos \delta\end{array}\right|$
$+\left|\begin{array}{ccc}
\sin \alpha & \cos \alpha & \cos \alpha \cdot \sin \delta \\
\sin \beta & \cos \beta & \cos \beta \cdot \sin \delta \\
\sin \gamma & \cos \gamma & \cos \gamma \cdot \sin \delta
\end{array}\right|$
$=\cos \delta\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin \alpha \\ \sin \beta & \cos \beta & \sin \beta \\ \sin \gamma & \cos \gamma & \sin \gamma\end{array}\right|$
$+\sin \delta\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \cos \alpha \\ \sin \beta & \cos \beta & \cos \beta \\ \sin \gamma & \cos \gamma & \cos \gamma\end{array}\right|$
$$
=\cos \delta \times 0+\sin \delta \times 0
$$
$\left(\because C_{1}\right.$ and $C_{2}$ are identical $=0$ $C_{2}$ and $C_{3}$ identical.)
Hence, the correct answer is option: 'A'.