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

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(C) Given equations are $\cos x + \cos y = -\cos \alpha$ and $\sin x + \sin y = -\sin \alpha$. Using the sum-to-product formulas,we have: $2 \cos \lef

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$\cot \alpha $

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(C) Given equations are $\cos x + \cos y = -\cos \alpha$ and $\sin x + \sin y = -\sin \alpha$. Using the sum-to-product formulas,we have: $2 \cos \left( \frac{x + y}{2} ight) \cos \left( \frac{x - y}{2} ight) = -\cos \alpha$ $(i)$ $2 \sin \left( \frac{x + y}{2} ight) \cos \left( \frac{x - y}{2} ight) = -\sin \alpha$ $(ii)$ Dividing equation $(ii)$ by equation $(i)$,we get: $\frac{2 \sin \left( \frac{x + y}{2} ight) \cos \left( \frac{x - y}{2} ight)}{2 \cos \left( \frac{x + y}{2} ight) \cos \left( \frac{x - y}{2} ight)} = \frac{-\sin \alpha}{-\cos \alpha}$ $	an \left( \frac{x + y}{2} ight) = 	an \alpha$ Taking the reciprocal on both sides: $\cot \left( \frac{x + y}{2} ight) = \cot \alpha$.

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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