If cos x + cos y + cos alpha = 0 and sin x + | vedclass

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(C) Given equations are $\cos x + \cos y = -\cos \alpha$ $(i)$ and $\sin x + \sin y = -\sin \alpha$ $(ii)$.Using the sum-to-product formulas:$\cos x +

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$\cot \alpha $

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(C) Given equations are $\cos x + \cos y = -\cos \alpha$ $(i)$ and $\sin x + \sin y = -\sin \alpha$ $(ii)$.Using the sum-to-product formulas:$\cos x + \cos y = 2 \cos \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right) = -\cos \alpha$ $(iii)$$\sin x + \sin y = 2 \sin \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right) = -\sin \alpha$ $(iv)$Dividing equation $(iii)$ by equation $(iv)$:$\frac{2 \cos \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right)}{2 \sin \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right)} = \frac{-\cos \alpha}{-\sin \alpha}$$\cot \left( \frac{x+y}{2} \right) = \cot \alpha$.

If $\cos x + \cos y + \cos \alpha = 0$ and $\sin x + \sin y + \sin \alpha = 0,$ then $\cot \left( \frac{x + y}{2} \right) = $

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