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 answer
    A
    0
  • B
    1
  • C
    $1+\sin \alpha \sin \beta \sin \gamma$
  • D
    $1-(\sin \alpha-\sin \beta)(\sin \beta-\sin \gamma)$ $(\sin \gamma-\sin \alpha)$

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'.

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