Question

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

• A. $\sin \alpha$
• B. $\cos \alpha$
• C. $\cot \alpha$
• D. $2$

Answer 1

Answer: (C)

$\cos x+\cos y+\cos \alpha =0$

$\cos x+\cos y=-\cos \alpha$

$[\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)]$

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

$\sin x+\sin y+\sin \alpha =0$

$\sin x+\sin y=-\sin \alpha$

$[\sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)]$

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

divide (i) by (ii)

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

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