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Question

In a ΔABC, prove that a(cos Ccos B)=2(bc)cos2A2.

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Solution

Let a=k sin Aa(cos Ccos B)=2(bc)cos2A2LHS=a(cos Ccos B)=a2.sinC+B2.sinBC2=2k sin A sinπA2.sinBC2=2k2 sinA2.cosA2.cosA2.sinBC2=2K2 cos2A2(2sinBC2.sinA2)=2k cos2A2(2sinBC2.sinπ(BC)2)=2K cos2A2(sin Bsin C)=2 cos2A2 (k sin Bk sin C)=2 cos2A2 (bc)=RHS.


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