$|\begin{array}{lll} \cos (θ + α) \sin (θ + α) 1 \\ \cos (θ + β) \sin (θ + β) 1 \\ \cos (θ + γ) \sin (θ + γ) 1 \end{array}|$ is independent of

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$α$

$β$

$γ$

$θ$

answer is D.

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Detailed Solution

Differentiating the given determinant w.r.t. $θ$

we get

$\begin{aligned} f^{'} (θ) = |\begin{array}{lll} - \sin (θ + α) & \sin (θ + α) & 1 \\ - \sin (θ + β) & \sin (θ + β) & 1 \\ - \sin (θ + γ) & \sin (θ + γ) & 1 \end{array}| \\ + |\begin{array}{lll} \cos (θ + α) & \cos (θ + α) & 1 \\ \cos (θ + β) & \cos (θ + β) & 1 \\ \cos (θ + γ) & \cos (θ + γ) & 1 \end{array}| \\ = 0 + 0 = 0 \end{aligned}$

where f(q) denotes the given determinant fi f(q) is independent of q. Alternately expanding the determinant along last column

we get $\begin{aligned} \sin (θ + γ - θ - β) - \sin (θ + γ - θ - α) + \sin (θ + β - θ - α) \\ = \sin (γ - β) + \sin (β - α) + \sin (α - γ) \end{aligned}$