Question

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

• A. $\sin \alpha$
• B. $\cos \alpha$
• C. $\tan \alpha$
• D. $\cot \alpha$

Answer 1

The correct option is D $\cot \alpha$

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

$=\frac{2\cos \frac{x+y}{2}\cos \frac{x-y}{2}}{2\sin \frac{x+y}{2}\cos \frac{x-y}{2}}$

$=\frac{\cos x+\cos y}{\sin x+\sin y}$

Now, since $\cos x+\cos y+\cos \alpha =0$ and $\sin x+\sin y+\sin \alpha =0$

$\Rightarrow \cos x+\cos y=-\cos \alpha$ and $\sin x+\sin y=-\sin \alpha$

Hence the given expression reduces to

$=\frac{-\cos \alpha }{-\sin \alpha }$

$=\cot \alpha$