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

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(B) Given equations are: $\cos x + \cos y = -\cos \alpha$  ... $(1)$ $\sin x + \sin y = -\sin \alpha$  ... $(2)$ Using sum-to-product formulas: $2 \co

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$\cot \alpha$

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(B) Given equations are: $\cos x + \cos y = -\cos \alpha$  ... $(1)$ $\sin x + \sin y = -\sin \alpha$  ... $(2)$ Using sum-to-product formulas: $2 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = -\cos \alpha$  ... $(3)$ $2 \sin \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = -\sin \alpha$  ... $(4)$ Dividing equation $(4)$ by equation $(3)$: $\frac{2 \sin \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight)}{2 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight)} = \frac{-\sin \alpha}{-\cos \alpha}$ $	an \left(\frac{x+y}{2}ight) = 	an \alpha$ Therefore,$\cot \left(\frac{x+y}{2}ight) = \cot \alpha$.

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

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