Question

prove: (cos(x) - sin(y))/(cos(y) - sin(x))=(cos(y) + sin(x))/(cos(x) + sin(y))

Answer 1

\[\LARGE \frac{(cosx-siny)}{cosy-sinx}=\frac{(cosy+sinx)}{cosx+siny}\]
Is this the question?

Answer 2

yes :)

Answer 3

look at the equation and see where you need to end up... starting from the left, it looks like you need to end up with a (cosx + siny) in your denominator... this is what i'd start with:
\(\large \frac{cosx-siny}{cosy-sinx}=\frac{cosx-siny}{cosy-sinx} \cdot \frac{cosx+siny}{cosx+siny}= \)...
try continue from here...
tag me if you need more help...

Answer 4

@lesliesaree

Welcome to OpenStudy. I apologize for the inappropriate comment by @you. I hope you will give us another chance to show that we can be and want to be of help to you.

Answer 5

\[\newcommand{cx}{\cos x}\newcommand {cy}{\cos y}\newcommand{sx}{\sin x}\newcommand{sy}{\sin y}\\\Large{\frac{\cx - \sy}{\cy - \sx}=\frac{\cy+\sx}{\cy+\sx}\cdot\frac{\cx - \sy}{\cy - \sx}\cdot\frac{\cx+\sy}{\cx+\sy}\\\quad=\frac{(\cy+\sx)\{(\cx-\sy)(\cx+\sy)\}}{\{(\cy+\sx)(\cy-\sx)\}(\cx+\sy)}\\\quad=\frac{(\cy+\sx)\{\cos^2x-\sin^2y\}}{\{\cos^2y-\sin^2x\}(\cx+\sy)}\\\quad =\frac{(\cy+\sx)\{1-\sin^2x-1+\cos^2y\}}{\{\cos^2y-\sin^2x\}(\cx+\sy)}\\\quad=\frac{(\cy+\sx)\{\cos^2y-\sin^2x\}}{\{\cos^2-\sin^2x\}(\cx+\sy)}\\\quad=\frac{\cy+\sx}{\cx+\sy}}\]

Answer 6

thank you so much for helping me @sirm3d and @dpaInc ! & I'm glad to know @Directrix I wasn't too sure at first, but I am so glad to find a resource like this! :)

Answer 7

Thanks for staying with us.