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Question

In any  ΔABC, prove that

a(cos Ccos B)=2(bc)cos2A2.


Solution

Applying sine rule and cosine formula, we have

LHS = a(cos Ccos B)

 =  a[a2+b2c22ab(a2+c2b2)2ac]=(a2c+b2cc3a2bbc2+b2)2bc

(b3c3)+(b2cbc2)(a2ba2c)2bc=(b3c3)+bc(bc)a2(bc)2bc

= (bc)[(b2+c2+bc)+bca2]2bc=(bc).{(b2+c2a2)2bc+2bc2bc}

(bc){(b2+c2a2)2bc+1}=(bc)(1+cos A)

2(bc)cos2A2=RHS

 a(cos Ccos B)=2(bc)cos2A2

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