# Thread: Trig Identity - I'm so close (i think..)

1. ## Trig Identity - I'm so close (i think..)

Im still trying to figure out these identities. Ive never had so much trouble with math, ugh. I think I almost figured this one out but I must have taken a wrong turn somewhere. Here is the identity.

$\displaystyle \frac{1-\sin\theta}{\cos\theta}+\frac{\cos\theta}{1-\sin\theta}=2\sec\theta$

I started with the left side and used the lcm to get...

$\displaystyle \frac{1-\sin\theta-\sin\theta+\sin^2\theta+\cos^2\theta}{(1-\sin\theta)(\cos\theta)}$

Then simplified to get...

$\displaystyle \frac{2-2\sin\theta}{(1-\sin\theta)(\cos\theta)}$

and finally...

$\displaystyle \frac{1-\sin\theta}{\cos\theta}$

I obviously messed something up somewhere can anyone tell me where I went wrong?

2. Originally Posted by repete
Im still trying to figure out these identities. Ive never had so much trouble with math, ugh. I think I almost figured this one out but I must have taken a wrong turn somewhere. Here is the identity.

$\displaystyle \frac{1-\sin\theta}{\cos\theta}+\frac{\cos\theta}{1-\sin\theta}=2\sec\theta$

I started with the left side and used the lcm to get...

$\displaystyle \frac{1-\sin\theta-\sin\theta+\sin^2\theta+\cos^2\theta}{(1-\sin\theta)(\cos\theta)}$

Then simplified to get...

$\displaystyle \frac{2-2\sin\theta}{(1-\sin\theta)(\cos\theta)}$

e^(i*pi) - you're fine up to here

and finally...

$\displaystyle \frac{1-\sin\theta}{\cos\theta}$

I obviously messed something up somewhere can anyone tell me where I went wrong?
$\displaystyle \frac{2-2\sin\theta}{(1-\sin\theta)(\cos\theta)} = \frac{2(1-\sin\theta)}{(1-\sin\theta)(\cos\theta)}$

$\displaystyle 1-sin\theta$ will cancel leaving $\displaystyle \frac{2}{cos\theta} = 2sec\theta$

3. Oh! Awesome Thanks!