1. ## Verifying Trig Identities

I understand the concept and know the identities, however our teacher offered us more difficult bonus problems and this one threw me for a loop.

The problem is

the square root of (1-sin(theta))/(1+sin(theta)) = 1/(sec(thetha) - tan(theta))

My question is... can you square both sides? I tried that and worked the right side until I got (1-sin^2(theta))/(1+sin^2(theta)) ... which, if I took the square root of, I would get (1-sin(theta))/(1+sin(theta)) however I can't just take the square root of one side.

I really don't know where to go with this problem..

Any help is appreciated!

Also, sorry if it makes no sense when I wrote it out.. if you would like me to write it out as an image so it makes more sense, I can. Thanks!

2. Originally Posted by mgannon93
I understand the concept and know the identities, however our teacher offered us more difficult bonus problems and this one threw me for a loop.

The problem is

the square root of (1-sin(theta))/(1+sin(theta)) = 1/(sec(thetha) - tan(theta))

My question is... can you square both sides? I tried that and worked the right side until I got (1-sin^2(theta))/(1+sin^2(theta)) ... which, if I took the square root of, I would get (1-sin(theta))/(1+sin(theta)) however I can't just take the square root of one side.

I really don't know where to go with this problem..

Any help is appreciated!

Also, sorry if it makes no sense when I wrote it out.. if you would like me to write it out as an image so it makes more sense, I can. Thanks!
Please, check if it is 1/ there.

3. Here it is typed out. (Hopefully this works, I'm posting it as an image).
Does this make more sense?

4. Originally Posted by mgannon93
Here it is typed out. (Hopefully this works, I'm posting it as an image).
Does this make more sense?

AHHHH1!!! It's +tan x!!!!
OK.
multiply the numerator and the denominator on the left side by (1-sinx)
ROOT of fraction
(1-sinx)(1-sinx)/ 1/
(1+sinx)(1-sinx) = 1/ sinx/
cosx + cosx

ROOT
numerator(1-sin)^2 / = 1/
denominator(1^2-sin^2x) (1+sinx)/
cosx

by sin^2x+cos^2x=1(Pythagorean Identity) denom is cos^2x
taking root we have on the left
(1-sinx)/cosx

On the right: 1 divide by fraction (1+sinx)/cosx is 1multiply by reciprocal cosx/(1+sinx). So it's cosx/(1+sinx).
We've got a proportion:
(1-sinx) cosx
cosx = (1+sinx)

Cross multiply we get (1-sinx)(1+sinx)=cos^2x, open parenthesses and we get 1-sin^2x=cos^2x(Pythogorean Identity)

5. Hello, mgannon93!

$\sqrt{\frac{1-\sin\theta}{1+\sin\theta}} \;=\;\frac{1}{\sec\theta - \tan\theta}$

$\sqrt{\frac{1-\sin\theta}{1+\sin\theta}\cdot{\color{blue}\frac{1 +\sin\theta}{1+\sin\theta}}} \;=\;\sqrt{\frac{1-\sin^2\!\theta}{(1+\sin\theta)^2}} \;=\;\sqrt{\frac{\cos^2\!\theta}{(1+\sin\theta)^2} }$ . $= \;\frac{\cos\theta}{1 + \sin\theta}$

Divide top and bottom by $\cos\theta\!:\quad \frac{\dfrac{\cos\theta}{\cos\theta}}{\dfrac{1}{\c os\theta} + \dfrac{\sin\theta}{\cos\theta}} \;=\;\frac{1}{\sec\theta + \tan\theta}

$

6. Thank you both so much!

I get it now.

Thanks!