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

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(D) Given,$\cos x+\cos y=-\cos \alpha$  $(i)$ Using the formula $\cos C + \cos D = 2 \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$,

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$\cot \alpha$

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(D) Given,$\cos x+\cos y=-\cos \alpha$  $(i)$ Using the formula $\cos C + \cos D = 2 \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$,we get: $2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = -\cos \alpha$ Also,$\sin x+\sin y=-\sin \alpha$  (ii) Using the formula $\sin C + \sin D = 2 \sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$,we get: $2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = -\sin \alpha$ Dividing equation $(i)$ by equation (ii): $\frac{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)} = \frac{-\cos \alpha}{-\sin \alpha}$ $\cot \left(\frac{x+y}{2}\right) = \cot \alpha$

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

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