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Question no. 15. In a Triangle ABC,summation CosA/SinBSinC =

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2 years ago

```							cosA/(sinBsinC) + cosB/(sinCsinA) + cosC/(sinAsinB)=2 2 (sinBsinC) = cos ( (B – C)/2) -cos ( (B + C)/2) cosA/(sinBsinC) + cosB/(sinCsinA) + cosC/(sinAsinB = (sin A cos A + sin B cos B + sin C cos C|) / (sin A sin B sin C) = (sin (2A) + sin (2B) + sin (2C) ) / (2 sin A sin B sin C) = (2 sin (A + B) cos (A – B) + 2 sin C cos C) / (2 sin A sin B sin C) = (sin C cos (A – B) – sin C cos (A + B)) / (sin A sin B sin C), since A + B = π – C = (cos (A – B) – cos (A + B) ) / (sin A sin B) = 2 (sin A sin B) / (sin A sin B), by the standard expansion, = 2
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2 years ago
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