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

  1. #1
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    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.

    \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...

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

    Then simplified to get...

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

    and finally...

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

    I obviously messed something up somewhere can anyone tell me where I went wrong?
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by repete View Post
    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.

    \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...

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

    Then simplified to get...

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

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

    and finally...

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

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

    1-sin\theta will cancel leaving \frac{2}{cos\theta} = 2sec\theta
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  3. #3
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    Oh! Awesome Thanks!
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