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  4. If cos x+ cos y+ cos α=0 and sin x+ sin y+ sin α=0 then cot ((x+y/2))=

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Q. If $\cos x+\cos y+\cos \alpha=0$ and $\sin x+\sin y+\sin \alpha=0$ then $\cot \left(\frac{x+y}{2}\right)=$

Trigonometric Functions

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
$\frac{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}=\frac{\cos \alpha}{\sin \alpha}$
$\Rightarrow \cot \left(\frac{x+y}{2}\right)=\cot \alpha$