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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)$

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$$\begin{align} \frac{\cos{x}}{1-\tan{x}}+\frac{\sin{x}}{1-\cot{x}} &=\frac{\cos{x}}{1-\tan{x}}\cdot\frac{\cos{x}}{\cos{x}}+\frac{\sin{x}}{1-\cot{x}}\cdot\frac{\sin{x}}{\sin{x}}\\ &=\frac{\cos^2{x}}{\cos{x}-\sin{x}}+\frac{\sin^2{x}}{\sin{x}-\cos{x}}\\ &=\frac{\cos^2{x}}{\cos{x}-\sin{x}}-\frac{\sin^2{x}}{\cos{x}-\sin{x}}\\ &=\frac{\cos^2{x}-\sin^2{x}}{\cos{x}-\sin{x}}\\ &=\frac{(\cos{x}-\sin{x})(\cos{x}+\sin{x})}{\cos{x}-\sin{x}}\\ &=\cos{x}+\sin{x}. \end{align}$$

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  • $\begingroup$ I didn't see it, omg, that's neet. $\endgroup$
    – Luther
    Jul 14 '14 at 13:11
  • $\begingroup$ providing the identity of "1-tanx=1-cotx" could have solved it? even though it seems impossible? $\endgroup$
    – Luther
    Jul 14 '14 at 13:18
  • $\begingroup$ @Luther That "identity" is false, and even if it were true it still wouldn't help. $\endgroup$
    – David H
    Jul 14 '14 at 13:23
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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 $$\frac{x}{1-\frac{y}{x}}+\frac{y}{1-\frac{x}{y}}=y+x\ .$$

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  • $\begingroup$ looking at trig equations in terms of x and y is great help $\endgroup$
    – Luther
    Jul 14 '14 at 14:19

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