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. $\tan \alpha$
• D. $\cot \alpha$

Answer 1

The correct option is D $\cot \alpha$

$cosx+cosy+cos\alpha =0$

$⟹cosx+cosy=-cos\alpha$

$⟹2cos\left(\frac{x+y}{2}\right)cos\left(\frac{x-y}{2}\right)=-cos\alpha$ ...(1)

$sinx+siny+sin\alpha =0$

$⟹sinx+siny=-sin\alpha$

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

Dividing equation (1) by (2), we get

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

$⟹cot\left(\frac{x+y}{2}\right)=cot\alpha$

Hence, answer is option (D)