A) \[\sin \alpha \]
B) \[\cos \alpha \]
C) \[\cot \alpha \]
D) \[\sin \,\left( \frac{x+y}{2} \right)\]
Correct Answer: C
Solution :
Given equation \[\cos x+\cos y+\cos \alpha =0\] and \[\sin x+\sin y+\sin \alpha =0.\] The given equation may be written as \[\cos x+\cos y=-\cos \alpha \] and \[\sin x+\sin y=-\sin \alpha .\]Therefore \[2\cos \left( \frac{x+y}{2} \right)\cos \left( \frac{x-y}{2} \right)=-\cos \alpha \] ?..(i) \[2\sin \left( \frac{x+y}{2} \right)\cos \left( \frac{x-y}{2} \right)=-\sin \alpha \] ?..(ii) Divide (i) by (ii), we get \[\frac{2\cos \left( \frac{x+y}{2} \right)\cos \left( \frac{x-y}{2} \right)}{2\sin \left( \frac{x+y}{2} \right)\cos \left( \frac{x-y}{2} \right)}\] \[=\frac{\cos \alpha }{\sin \alpha }\] Þ \[\cot \left( \frac{x+y}{2} \right)=\cot \alpha \].
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