Question

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

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

Answer 1

The correct option is C $\cot \theta$

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

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

$\frac{2\cos (x+y)}{2}\times \frac{\cos (x-y)}{2}=-\cos \theta ⟶1$

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

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

$\frac{2\sin (x+y)}{2}\times \frac{\sin (x-y)}{2}=-\sin \theta$

Dividing both,

$\frac{\frac{2\cos (x+y)}{2}\times \frac{\cos (x-y)}{2}}{\frac{2\sin (x+y)}{2}\times \frac{\sin (x-y)}{2}}=\frac{-\cos \theta }{-\sin \theta }$

$\frac{\cot (x+y)}{2}=\cot \theta$