If A = begin{vmatrix} sin(theta + alpha) & co | vedclass

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(D) Given $A = \begin{vmatrix} \sin(	heta + \alpha) & \cos(	heta + \alpha) & 1 \ \sin(	heta + \beta) & \cos(	heta + \beta) & 1 \ \sin(	heta + \

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$A$ is independent of $	heta$

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(D) Given $A = \begin{vmatrix} \sin(	heta + \alpha) & \cos(	heta + \alpha) & 1 \ \sin(	heta + \beta) & \cos(	heta + \beta) & 1 \ \sin(	heta + \gamma) & \cos(	heta + \gamma) & 1 \end{vmatrix}$.Applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$A = \begin{vmatrix} \sin(	heta + \alpha) & \cos(	heta + \alpha) & 1 \ \sin(	heta + \beta) - \sin(	heta + \alpha) & \cos(	heta + \beta) - \cos(	heta + \alpha) & 0 \ \sin(	heta + \gamma) - \sin(	heta + \alpha) & \cos(	heta + \gamma) - \cos(	heta + \alpha) & 0 \end{vmatrix}$.Expanding along the third column:$A = 1 \cdot [(\sin(	heta + \beta) - \sin(	heta + \alpha))(\cos(	heta + \gamma) - \cos(	heta + \alpha)) - (\cos(	heta + \beta) - \cos(	heta + \alpha))(\sin(	heta + \gamma) - \sin(	heta + \alpha))]$.Using the identity $\sin(x)\cos(y) - \cos(x)\sin(y) = \sin(x-y)$:$A = \sin((	heta + \beta) - (	heta + \gamma)) + \sin((	heta + \alpha) - (	heta + \beta)) + \sin((	heta + \gamma) - (	heta + \alpha))$.$A = \sin(\beta - \gamma) + \sin(\alpha - \beta) + \sin(\gamma - \alpha)$.Since the result contains only constants $\alpha, \beta, \gamma$,$A$ is independent of $	heta$.

If $A = \begin{vmatrix} \sin(\theta + \alpha) & \cos(\theta + \alpha) & 1 \\ \sin(\theta + \beta) & \cos(\theta + \beta) & 1 \\ \sin(\theta + \gamma) & \cos(\theta + \gamma) & 1 \end{vmatrix}$,then

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