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.

  • 3
    $\begingroup$ I think you're missing some absolute value signs... $\endgroup$ – Micah Dec 3 '14 at 19:40
  • 1
    $\begingroup$ @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

Hint: Multiply the quantity under the root by $$\frac{1-\sin(x)}{1-\sin(x)}$$


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





$\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.

  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.