A) \[\sin \alpha \]
B) \[\cos \alpha \]
C) \[\cot \alpha \]
D) \[\sin \left( \frac{x+y}{2} \right)\]
Correct Answer: C
Solution :
\[\cos x+\cos y=-\cos \alpha \] ?(i) \[\sin x+\sin y=-\sin \alpha \] ?(ii) Divide equation (i) by (ii) we get \[\frac{\cos x+\cos y}{\sin x+\sin y}=\frac{\cos \alpha }{\sin \alpha }\] \[\Rightarrow \]\[\frac{2\cos \frac{x+y}{2}\cos \frac{x-y}{2}}{2\sin \frac{x+y}{2}\cos \frac{x-y}{2}}=\cot \alpha \] \[\Rightarrow \]\[\cot \frac{x+y}{2}=\cot \alpha \]
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