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

Oct 2009
12
0
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?
 

e^(i*pi)

MHF Hall of Honor
Feb 2009
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West Midlands, England
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 \(\displaystyle \frac{2}{cos\theta} = 2sec\theta\)
 
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