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Question: If \(\cos x + \cos y + \cos\alpha = 0\) and \(\sin x + \sin y + \sin\alpha = 0,\) then \(\cot\left( ...

If cos⁡x+cos⁡y+cos⁡α=0\cos x + \cos y + \cos\alpha = 0 and sin⁡x+sin⁡y+sin⁡α=0,\sin x + \sin y + \sin\alpha = 0, then cot⁡(x+y2)=\cot\left( \frac{x + y}{2} \right) =

A

sin⁡α\sin\alpha

B

cos⁡α\cos\alpha

C

cot⁡α\cot\alpha

D

sin⁡(x+y2)\sin\left( \frac{x + y}{2} \right)

Answer

cot⁡α\cot\alpha

Explanation

Solution

Given equation cos⁡x+cos⁡y+cos⁡α=0\cos x + \cos y + \cos\alpha = 0 and

sin⁡x+sin⁡y+sin⁡α=0.\sin x + \sin y + \sin\alpha = 0. The given equation may be written a

cos⁡x+cos⁡y=−cos⁡α\cos x + \cos y = - \cos\alpha and sin⁡x+sin⁡y=−sin⁡α.\sin x + \sin y = - \sin\alpha. Therefore

2cos⁡(x+y2)cos⁡(x−y2)=−cos⁡α2\cos\left( \frac{x + y}{2} \right)\cos\left( \frac{x - y}{2} \right) = - \cos\alpha …..(i)

2sin⁡(x+y2)cos⁡(x−y2)=−sin⁡α2\sin\left( \frac{x + y}{2} \right)\cos\left( \frac{x - y}{2} \right) = - \sin\alpha …..(ii)

Divide (i) by (ii), we get 2cos⁡(x+y2)cos⁡(x−y2)2sin⁡(x+y2)cos⁡(x−y2)\frac{2\cos\left( \frac{x + y}{2} \right)\cos\left( \frac{x - y}{2} \right)}{2\sin\left( \frac{x + y}{2} \right)\cos\left( \frac{x - y}{2} \right)}

=cos⁡αsin⁡α= \frac{\cos\alpha}{\sin\alpha} ⇒ cot⁡(x+y2)=cot⁡α\cot\left( \frac{x + y}{2} \right) = \cot\alpha.