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Can somebody help to prove that:

$$\frac{\cos a - \sin a + 1}{\cos a + \sin a - 1 } = \frac{\sin a}{1-\cos a}$$

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marked as duplicate by Carl Mummert, lab bhattacharjee trigonometry Feb 5 '18 at 4:28

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    $\begingroup$ Please show some efforts first. $\endgroup$ – Saad Feb 5 '18 at 2:54
  • $\begingroup$ Multiply by both denominators and cancel out terms. There would be left $1=\sin^2 a+\cos^2 a$. $\endgroup$ – user_194421 Feb 5 '18 at 3:01
  • $\begingroup$ Multiply thru by the product of the denominators and simplify. I.e. if $B\ne 0\ne D$ then $A/B=C/D\iff AD=BC.$ $\endgroup$ – DanielWainfleet Feb 5 '18 at 3:02
  • $\begingroup$ As requested by Alex, it's generally best for questioners to include a sense of what they know about a problem and where they got stuck, so that potential answerers can target their responses to an appropriate skill level, and without wasting time telling the questioner things that are already known. Posting a problem without this kind of context looks like you're just trying to get people to do your homework for you, often resulting in down-votes and/or votes-to-close. $\endgroup$ – Blue Feb 5 '18 at 3:04
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\begin{eqnarray*} \frac{ \cos a -\sin a +1}{\cos a +\sin a -1} \times \frac{1-\cos a}{1-\cos a} =\frac{\color{red}{\cos a}-\sin a+\overbrace{1-\cos^2 a}^{\sin^2 a}+\sin a \cos a\color{red}{-\cos a}}{(\cos a +\sin a -1)((1-\cos a)} =\frac{\sin a (\cos a +\sin a -1)}{(1-\cos a)(\cos a +\sin a -1)} \end{eqnarray*}

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  • $\begingroup$ Thanks man, it really helped $\endgroup$ – user140202 Feb 5 '18 at 3:26

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