-1
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$x=\cfrac{2\sin A}{1+\cos A+\sin A}$ , then $ \cfrac{1-\cos A- \sin A}{\cos A}=?$

Options are

$1$. $x$

$2$. $\cfrac 1x$

$3$. $-x$

$4$. $\cfrac{-1}x$

Can not attend it.

I don't know how to approach it.

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closed as off-topic by user296602, GNUSupporter 8964民主女神 地下教會, Matthew Conroy, man and laptop, Namaste Feb 8 '18 at 0:42

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If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ maybe you can try the different options and see the one who leads to tautologic (obvious) equality ? (If you really don't have another idea) $\endgroup$ – Netchaiev Feb 7 '18 at 18:18
  • $\begingroup$ Try en.m.wikipedia.org/wiki/… $\endgroup$ – lab bhattacharjee Feb 7 '18 at 18:42
  • $\begingroup$ laliga (account number ?) This site is not a homework completion service, despite, like we see below, there are users so desperate for rep, that they'll answer anything. You posted what isn't even a question. It is a transcription of a problem assigned to you, for you to do. You failed to even attempt it. If you "don't know how to approach it", it's time to read your text and lecture notes which cover the material you failed to read, or attend to. Don't come here and hand us your homework. You are welcome to post here if and only if you're willing to help us help you. $\endgroup$ – Namaste Feb 8 '18 at 0:46
  • $\begingroup$ this is not my homework and i am in class 6 $\endgroup$ – laliga Feb 8 '18 at 18:28
2
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Let $$ y ={\cos A \over 1-\cos A- \sin A}$$ We are interested in ${1\over y}$. Then $$xy = {2\sin A\cos A \over 1-(\cos A+\sin A)^2 } ={\sin 2A \over 1-\cos ^2A -\sin^2 A -2\sin A \cos A } =- 1$$

So ${1\over y} = -x$

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0
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$x=\dfrac{2sinA}{1+cosA+sinA}.\left(\dfrac{1-cosA-sinA}{1-cosA-sinA}\right)$

$x=\dfrac{{2sinA}.({1-cosA-sinA})}{1-(cosA+SinA)^{2}}=-\dfrac{({1-cosA-sinA})}{cosA}$

$\implies\dfrac{({1-cosA-sinA})}{cosA}=-x$

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0
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$$(\cos A+\sin A+1)(\cos A+\sin A-1)=\cdots=2\sin A\cos A$$

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