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In triangle ABC, prove that a(cosC-cosB...

In `triangle ABC`, prove that `a(cosC-cosB)=2(b-c)cos^2.(A)/(2)` .

Text Solution

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By projection rule ,
`b=a cosC+c cos A` and `c=acos B +bcos A `
`therefore b-c=(acosC+c cos A)-(acos B +bcosA)=a(cos C-cosB )-(b-c)cosA`
`therefore (b-c)+(b-c)cosA=a(cos C-cos B)`
`therefore (b-c)(1+cosA)=a(cos C-cos B)`
`therefore (b-c)xx2cos^2.(A)/(2)=a(cos C -cos B)`
`therefore a(cos C-cos B)=2(b-c)cos^2.(A)/(2)`.
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