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Prove $\sqrt\frac{1 - \sin x}{1 + \sin x} = \frac{1}{\cos x} - \tan x$

I tried but I couldn't figure it out, give me a hint please.

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    $\begingroup$ I think you're missing some absolute value signs... $\endgroup$ – Micah Dec 3 '14 at 19:40
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    $\begingroup$ @Micah I just copy & paste from textbook $\endgroup$ – omidh Dec 3 '14 at 19:45
  • $\begingroup$ This is a basic level trigonometry question. I think it doesn't have approach to going beyond 90 degree angle. and less than 0 degree angle. $\endgroup$ – Hardey Pandya Dec 3 '14 at 20:17
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Hint: Multiply the quantity under the root by $$\frac{1-\sin(x)}{1-\sin(x)}$$

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It is wrong, it isn't the same, if you want it for every value!! Because:

$$\frac{1}{cos(x)}-tan(x)=sec(x)-tan(x)$$

and:

$$(\frac{1-sin(x)}{1+sin(x)})^{\frac{1}{2}}=(\frac{2}{sin(x)+1}-1)^{\frac{1}{2}}$$

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$\sqrt\frac{1 - \sin x}{1 + \sin x} $

=$\sqrt\frac{(1 - \sin x)(1 - \sin x)}{(1 - \sin x)(1 + \sin x)}$

=$\sqrt\frac{(1 - \sin x)^2}{(1 - \sin^2 x)}$

=$\sqrt\frac{(1 - \sin x)^2}{(\cos^2x)}$
[because $sin^2 x + \cos^2 x =1$]

=$\frac{1-\sin x}{\cos x}$

=$\frac{1}{cosx}-\frac{\sin x}{\cos x}$

=$\frac{1}{cosx}-\tan x$

NOTE : This answer holds only if $sine$ and $cosine$ functions are in first quadrant.

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    $\begingroup$ omidh asked for a hint, not an outright answer. Please respect their wishes, as they may have wanted to have the experience of figuring out the problem on their own. $\endgroup$ – graydad Dec 3 '14 at 19:47
  • $\begingroup$ @graydad agreed $\endgroup$ – homegrown Dec 3 '14 at 19:50
  • $\begingroup$ Ah..... this is a new experience in mathstackexchange.... 2 downvotes!!! :p, , well I accept my mistake.. I read only the title of question. $\endgroup$ – Hardey Pandya Dec 3 '14 at 20:15

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