If cos x + cos y + cos alpha = 0 and sin x + | vedclass

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Question

(c) Given equation $\cos x + \cos y + \cos \alpha = 0$ 

and $\sin x + \sin y + \sin \alpha = 0.$

The given equation may be written as $\cos

Vedclass · Accepted Answer

$\cot \alpha $

Vedclass · Answer

(c) Given equation $\cos x + \cos y + \cos \alpha = 0$ 

and $\sin x + \sin y + \sin \alpha = 0.$

The given equation may be written as $\cos x + \cos y = - \cos \alpha $ 

and $\sin x + \sin y = - \sin \alpha .$

Therefore $2\cos \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x - y}}{2}} \right) = - \cos \alpha $&hellip;..$(i) $

$2\sin \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x - y}}{2}} \right) = - \sin \alpha $&hellip;..$(ii)$

Divide $(i)$ by $(ii)$, we get 

$\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)}}$$ = \frac{{\cos \alpha }}{{\sin \alpha }}$

==> $\cot \left( {\frac{{x + y}}{2}} \right) = \cot \alpha $.

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

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