# Prove trigonometry identity for $\sin A+\cos A$

I’ve been struggling in proving this identity for hours (yes, shame on me), but I can’t see any light.

$\frac { \cos(A) }{ 1-\tan(A) } +\frac { \sin(A) }{ 1-\cot(A) } =\sin(A)+\cos(A)$

I've been using Pythagorean equations/identities, maybe I’m going in the wrong direction.

Please provide the steps or hints to prove this equality?

I've also thought that a way was to check for the LHS equality of the denominators, could this be a way, or is it algebraically wrong?

$1-\tan(A)=1-\cot(A)$

\begin{align} \frac{\cos{x}}{1-\tan{x}}+\frac{\sin{x}}{1-\cot{x}} &=\frac{\cos{x}}{1-\tan{x}}\cdot\frac{\cos{x}}{\cos{x}}+\frac{\sin{x}}{1-\cot{x}}\cdot\frac{\sin{x}}{\sin{x}}\\ &=\frac{\cos^2{x}}{\cos{x}-\sin{x}}+\frac{\sin^2{x}}{\sin{x}-\cos{x}}\\ &=\frac{\cos^2{x}}{\cos{x}-\sin{x}}-\frac{\sin^2{x}}{\cos{x}-\sin{x}}\\ &=\frac{\cos^2{x}-\sin^2{x}}{\cos{x}-\sin{x}}\\ &=\frac{(\cos{x}-\sin{x})(\cos{x}+\sin{x})}{\cos{x}-\sin{x}}\\ &=\cos{x}+\sin{x}. \end{align}
Hint. Write $\tan A$ and $\cot A$ in terms of $\sin A$ and $\cos A$, then simplify the fractions. Don't forget how to factorise a difference of two squares.
Comment. Other than using the definitions of $\tan A$ and $\cot A$, this problem really has nothing to do with trigonometry. It is just a matter of proving $$\frac{x}{1-\frac{y}{x}}+\frac{y}{1-\frac{x}{y}}=y+x\ .$$