Q117 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics

$\left|\begin{array}{ccc} \cos (\alpha+\beta) & -\sin (\alpha+\beta) & \cos 2 \beta \\ \sin \alpha & \cos \alpha & \sin \beta \\ -\cos \alpha & \sin \alpha & \cos \beta \end{array}\right| $ is independent of

  • Correct answer
    A
    $\alpha$
  • B
    $\alpha$ and $\beta$
  • C
    $\beta$
  • D
    Neither $\alpha$ nor $\beta$

Solution

Apply $C_1 \to C_1\cos\beta + C_2\sin\beta$ and $C_2 \to -C_1\sin\beta + C_2\cos\beta$:

$\Delta = \begin{vmatrix} \cos(\alpha+2\beta) & -\sin(\alpha+2\beta) & \cos2\beta \\ \sin(\alpha+\beta) & \cos(\alpha+\beta) & \sin\beta \\ -\cos(\alpha+\beta) & \sin(\alpha+\beta) & \cos\beta \end{vmatrix}$

Expanding along first row:

$\Delta = \cos(\alpha+2\beta)[\cos(\alpha+\beta)\cos\beta - \sin(\alpha+\beta)\sin\beta]$
$+ \sin(\alpha+2\beta)[\sin(\alpha+\beta)\cos\beta + \cos(\alpha+\beta)\sin\beta]$
$+ \cos2\beta[\sin^2(\alpha+\beta) + \cos^2(\alpha+\beta)]$

$= \cos(\alpha+2\beta)\cos(\alpha+2\beta) + \sin(\alpha+2\beta)\sin(\alpha+2\beta) + \cos2\beta$

$= \cos^2(\alpha+2\beta) + \sin^2(\alpha+2\beta) + \cos2\beta$

$= 1 + \cos2\beta$

Result depends on $\beta$ but is independent of $\alpha$.

Hence, the correct answer is option: 'A'.

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