TS EAMCET · Maths · Trigonometric Equations
If \(\cos x+\cos y+\cos \alpha=0\) and \(\sin x+\sin y+\sin \alpha=0\), then \(\cot \left(\frac{x+y}{2}\right)\) is equal to
- A \(\sin \alpha\)
- B \(\cos \alpha\)
- C \(\tan \alpha\)
- D \(\cot \alpha\)
Answer & Solution
Correct Answer
(D) \(\cot \alpha\)
Step-by-step Solution
Detailed explanation
Given, \(\cos x+\cos y+\cos \alpha=0\) \[ \begin{aligned} & \therefore \quad \cos x+\cos y=-\cos \alpha \\ & \Rightarrow \quad 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=-\cos \alpha \\ & \end{aligned} \] And \(\sin x+\sin y+\sin \alpha=0\)…
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