Question

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

• A. $\sin \alpha$
• B. $\cos \alpha$
• C. $\cot \alpha$
• D. $\sin \left(\frac{x+y}{2}\right)$

Answer 1

Answer: (C)

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