The determinant xsin⁡θcos⁡θ−sin⁡θ−x1cos⁡θ1x is

Let Δ=xsin⁡θcos⁡θ−sin⁡θ−x1cos⁡θ1x By expanding along first row, we getΔ=x−x11x−sin⁡θ−sin⁡θ1cos⁡θx+cos⁡θ−sin⁡θ−xcos⁡θ1=x−x2−1−sin⁡θ(−xsin⁡θ−cos⁡θ)+cos⁡θ(−sin⁡θ+xcos⁡θ)=−x3−x+xsin2⁡θ+sin⁡θcos⁡θ−sin⁡θcos⁡θ+xcos2⁡θ=−x3−x+xsin2⁡θ+cos2⁡θ∵sin2⁡θ+cos2⁡θ=1=−x3−x+x=−x3Hence A is independent of θ.

The determinant xsin⁡θcos⁡θ−sin⁡θ−x1cos⁡θ1x is

Let Δ=xsin⁡θcos⁡θ−sin⁡θ−x1cos⁡θ1x By expanding along first row, we getΔ=x−x11x−sin⁡θ−sin⁡θ1cos⁡θx+cos⁡θ−sin⁡θ−xcos⁡θ1=x−x2−1−sin⁡θ(−xsin⁡θ−cos⁡θ)+cos⁡θ(−sin⁡θ+xcos⁡θ)=−x3−x+xsin2⁡θ+sin⁡θcos⁡θ−sin⁡θcos⁡θ+xcos2⁡θ=−x3−x+xsin2⁡θ+cos2⁡θ∵sin2⁡θ+cos2⁡θ=1=−x3−x+x=−x3Hence A is independent of θ.

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The determinant $|\begin{array}{ccc} x & \sin θ & \cos θ \\ - \sin θ & - x & 1 \\ \cos θ & 1 & x \end{array}|$ is

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independent of x only

None of the above

independent of $θ$ only

independent of both $θ$ and x

answer is A.

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Detailed Solution

Let $Δ = |\begin{array}{ccc} x & \sin θ & \cos θ \\ - \sin θ & - x & 1 \\ \cos θ & 1 & x \end{array}|$ By expanding along first row, we get

$\begin{array}{l} Δ = x |\begin{array}{cc} - x & 1 \\ 1 & x \end{array}| - \sin θ |\begin{array}{cc} - \sin θ & 1 \\ \cos θ & x \end{array}| + \cos θ |\begin{array}{cc} - \sin θ & - x \\ \cos θ & 1 \end{array}| \\ = x (- x^{2} - 1) - \sin θ (- x \sin θ - \cos θ) + \cos θ (- \sin θ + x \cos θ) \\ = - x^{3} - x + x \sin^{2} θ + \sin θ \cos θ - \sin θ \cos θ + x \cos^{2} θ \\ = - x^{3} - x + x (\sin^{2} θ + \cos^{2} θ) [∵ \sin^{2} θ + \cos^{2} θ = 1] \\ = - x^{3} - x + x = - x^{3} \end{array}$