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

If $\cos x+\cos y=-\cos \alpha, \sin x+\sin y=-\sin \alpha$, then $\cot \left(\frac{x+y}{2}\right)=$
  1. $-\cot \propto$
  2. $\cot \propto$
  3. $-\tan \propto$
  4. $\tan \propto$

Solution

Given $\cos x+\cos y=-\cos \alpha$ and $\sin x+\sin y=-\sin \alpha$ $2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=-\cos \alpha$ ...(1) and $2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=-\sin \alpha$ ...(2) Divided equation (1) by equation (2) $\cot \left(\frac{x+y}{2}\right)=\cot \alpha$

Asked in: MHT CET 2020 (13 Oct Shift 2)

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