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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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Verified Answer
The correct answer is:
$\cot \alpha$
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
$\begin{aligned}
& 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=-\cos \alpha \\
& 2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=-\sin \alpha \\
& \frac{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)} \\
& =\frac{\cos \alpha}{\sin \alpha} \Rightarrow \cot \left(\frac{x+y}{2}\right)=\cot \alpha \\
&
\end{aligned}$
$\begin{aligned}
& 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=-\cos \alpha \\
& 2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=-\sin \alpha \\
& \frac{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)} \\
& =\frac{\cos \alpha}{\sin \alpha} \Rightarrow \cot \left(\frac{x+y}{2}\right)=\cot \alpha \\
&
\end{aligned}$
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