If \( \Large \cos x + \cos y + \cos \alpha = 0 \) and \( \Large \sin x + \sin y + \sin \alpha = 0,\) then \( \Large \cot \left(\frac{x+y}{2}\right) \) is equal to:
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A) \( \Large \sin \alpha \) |
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B) \( \Large \cos \alpha \) |
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C) \( \Large \cot \alpha \) |
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D) \( \Large \sin \left(\frac{x+y}{2}\right) \) |
Correct Answer:
| C) \( \Large \cot \alpha \) |
Given equation are \( \Large \cos x + \cos y + \cos \alpha = 0 \) and \( \Large \sin x + \sin y + \sin \alpha = 0 \). Then given equations may be written as
\( \Large \cos x + \cos y = - \cos \alpha \)
and \( \Large \sin x + \sin y = - \sin \alpha \)
=> \( \Large 2 \cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right) = - \cos \alpha \) ...(i)
and \( \Large 2 \sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right) = - \sin \alpha \) ...(ii)
From equations (i) and (ii) we get
\( \Large \frac{2 \cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right) }{2 \sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right) } = \frac{\cos \alpha }{\sin \alpha } \)
=> \( \Large \cot \left(\frac{x+y}{2}\right) = \cot \alpha \)
Part of solved Trigonometric ratio questions and answers : >> Elementary Mathematics >> Trigonometric ratio
