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किसी `DeltaABC` में , सिद्ध कीजिए कि `a(cosC-cosB)=2(b-c)cos^2.(A)/(2)`.

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बायाँ पक्ष `=a[cosC-cosB]`
`=a[((a^2+b^2-c^2))/(2ab)-((a^2+c^2-b^2))/(2ac)]`
`=(a^2c+b^2c-c^3-a^2b-bc^2+b^3)/(2bc)`
`=((b^3-c^3)+(b^2-c-bc^2)-(a^2b-a^2c))/(2bc)`
`=((b^3-c^3)+bc(b-c)-a^2(b-c))/(2bc)`
`=(b-c)([(b^2+c^2+bc)+bc-a^2])/(2bc)`
`=(b-c){((b^2+c^2-a^2))/(2bc)+(2bc)/(2bc)}=(b-c){((b^2+c^2-a^2))/(2bc)+1}`
`=(b-c)(1+cosA)`
`=2(b-c)cos^2.(A)/(2)=` दायाँ पक्ष

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