In any triangle `A B C` , prove that: `(1+cos(A-B)cos C)/(1+cos(A-C)cos B)=(a^2+b^2)/(a^2+c^2)`

Explanation

We know that in any `triangleABC, a=k sinA, b=ksinB, c=ksinC`
`therefore` LHS can be written as
`=(k^(2)(sin^(2)A+sin^(2)B))/(k^(2)(sin^(2)A+sin^(2)C)) = (1-cos^(2)A + sin^(2)B)/(1-cos^(2)A+sin^(2)C)=(1-(cos^(2)A-sin^(2)B))/(1-(cos^(2)A-sin^(2)C))`..........(i)
Since `cos2A = 2cos^(2)A-1`
`cos^(2)A = (1+cos2A)/(2)`
`rArr sin^(2)A = (1-cos2A)/(2)`
`therefore` (i) reduces to `(1-1/2[1+cos2A-1+cos2B])/(1-1/2[1+cos2A-1+cos2C])`
`=(1-cos(A-B)cos(A+B))/(1-cos(A-C)cos(180^(@)-B))`
`=(1+cos(A-B)cosC)/(1+cos(A-C)cosB)`= RHS
Hence Proved.