Question: prove that cosx/(sinx+cosy)+cosy/(siny-cosx)=cosx/(sinx-cosy)+cosy/(siny+cosy)
There's a typo in your question. The final denominator should be (siny+cosx) not (siny+cosy).
Assumption: That the equation is true for all values of x and y, i.e. that it is an identity
cosx/(sinx+cosy) + cosy/(siny-cosx) = cosx/(sinx-cosy) + cosy/(siny+cosx) -- rearrange
cosx{1/(sinx+cosy) - 1/(sinx-cosy)} = cosy{1/(siny+cosy) - 1/(siny-cosx)} -- rationalise the denominators
cosx{[(sinx - cosy) - (sinx + cosy)]/[(sinx+cosy)(sinx-cosy)]} = cosy{[(siny - cosx) - (siny + cosx)]/[(siny+cosy)(siny-cosx)]}
cosx{(-2cosy)/(sin^2x - cos^2y)} = cosy{(-2cosx)/(sin^2y - cos^2y)}
cosx{(-2cosy)/[(1 - cos^2x) - (1 - sin^2y)]} = cosy{(-2cosx)/(sin^2y - cos^2y)}
(-2.cosx.cosy)/[sin^2y - cos^2x] = (-2.cosx.cosy)/(sin^2y - cos^2y)
Since lhs is identical with rhs, then the above equation is true for all values of x and y. It is an identity.
Hence the Assumption is true
Q.E.D.