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If any triangle `A B C`, that:`(b^2-c^2)/(cosB+cosC)+(c^2-a^2)/(cosC+cosA)+(a^2-b^2)/(cosA+cosB)=0` |
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Answer» From sine law, we have, `a/sinA = b/sinB = c/sinC = k` `=> a = ksinA, b = ksin B, c = ksinC` Now, `L.H.S. = (b^2-c^2)/(cosB+cosC)+(c^2-a^2)/(cosC+cosA)+(a^2-b^2)/(cosA+cosB)` `= (k^2(sin^2B-sin^2C))/(cosB+cosC)+ (k^2(sin^2C-sin^2A))/(cosC+cosA)+ (k^2(sin^2A-sin^2B))/(cosA+cosB)` `= (k^2((1-cos^2B)-(1-cos^2C)))/(cosB+cosC)+ (k^2((1-cos^2C)-(1-cos^2A)))/(cosC+cosA)+ (k^2((1-cos^2A)-(1-cos^2B)))/(cosA+cosB)` `=k^2[(cos^2B-cos^2C)/(cosB+cosC)+(cos^2C-cos^2A)/(cosC+cosA)+(cos^2A-cos^2B)/(cosA+cosB)]` `=k^2[((cosB+cosC)(cosB-cosC))/(cosB+cosC)+((cosC+cosA)(cosC-cosA))/(cosC+cosA)+((cosA+cosB)(cosA-cosB))/(cosA+cosB)]` `=k^2[cosB-cosC+cosC-cosA+cosA-cosB]` `=k^2[0] = 0 = R.H.S.` |
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