# Verifying Trig Identities

• Nov 5th 2008, 03:59 PM
mgannon93
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!
• Nov 5th 2008, 04:44 PM
tdschool
Quote:

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.
• Nov 5th 2008, 04:54 PM
mgannon93
Here it is typed out. (Hopefully this works, I'm posting it as an image).
Does this make more sense?

http://i33.tinypic.com/e13reg.jpg
• Nov 5th 2008, 05:46 PM
tdschool
Quote:

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

http://i33.tinypic.com/e13reg.jpg

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)
• Nov 5th 2008, 06:40 PM
tdschool
• Nov 5th 2008, 07:14 PM
Soroban
Hello, mgannon93!

Quote:

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

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

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

• Nov 6th 2008, 02:35 AM
mgannon93
Thank you both so much!

I get it now.

Thanks! (Happy)