If A+B+C =π/2 and sinBsinC = -sinA then what is the value of cotA+ cotB+cotC ?

Another Explanation (5): ```html

দেওয়া আছে, \(A+B+C = \frac{\pi}{2}\) এবং \( \sin B \sin C = -\sin A \)।

যেহেতু \(A+B+C = \frac{\pi}{2}\), তাই \(A = \frac{\pi}{2} - (B+C)\)।

অতএব, \( \sin A = \sin\left(\frac{\pi}{2} - (B+C)\right) = \cos(B+C) \)।

সুতরাং, \( \sin B \sin C = -\cos(B+C) \)।

আমরা জানি, \( \cos(B+C) = \cos B \cos C - \sin B \sin C \)।

তাহলে, \( \sin B \sin C = -(\cos B \cos C - \sin B \sin C) \)।

সুতরাং, \( \sin B \sin C = -\cos B \cos C + \sin B \sin C \)।

অতএব, \( 2 \sin B \sin C = -(\cos B \cos C - \sin B \sin C) \)

\(\Rightarrow\) \( \cos B \cos C = 0 \)।

এখন, \(\cot A + \cot B + \cot C = \cot A + \frac{\cos B}{\sin B} + \frac{\cos C}{\sin C}\)

\(= \cot A + \frac{\cos B \sin C + \cos C \sin B}{\sin B \sin C}\)

\(= \cot A + \frac{\sin(B+C)}{\sin B \sin C}\)

যেহেতু \(A+B+C = \frac{\pi}{2}\), তাই \(B+C = \frac{\pi}{2} - A\)।

সুতরাং, \(\sin(B+C) = \sin\left(\frac{\pi}{2} - A\right) = \cos A\).

অতএব, \(\cot A + \cot B + \cot C = \cot A + \frac{\cos A}{\sin B \sin C}\).

আমরা জানি, \(\sin B \sin C = -\sin A\).

তাহলে, \(\cot A + \cot B + \cot C = \cot A + \frac{\cos A}{-\sin A} = \cot A - \cot A = 0\).

সুতরাং, \(\cot A + \cot B + \cot C = 0\)। 🎉

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