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. $2\sin \alpha$

Answer 1

Answer: (C)

$\cot \alpha$

If $\cos x+\cos y+\cos \alpha =0\dots \dots \left(i\right)$

$\sin x+\sin y+\sin \alpha =0\dots \dots \left(ii\right)$

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

Now,

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

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

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

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

By $\left(iii\right)÷\left(iv\right)$

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

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

The correct option is C $\cot \alpha$.