If cos x + cos y + cos α = 0 and sin x + sin y + sin α = 0, then cot ,left( {frac{{x + y}}{2}} rig

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3.Trigonometrical Ratios, Functions and Identities

medium

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

$\sin \alpha $

$\cos \alpha $

$\cot \alpha $

$\sin \,\left( {\frac{{x + y}}{2}} \right)$

Solution

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 $…..$(i) $

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

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