Question

prove that:cosx/sinx+cosy+cosy/siny-cosx=cosx/sinx-cosy+cosy/siny+cosx

Answer 1

Answer:

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)

Answer 2

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)