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Math Help - verifying identities...another one i'm stuck on.

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    verifying identities...another one i'm stuck on.

    tan+\frac{cos}{1+sin} = sec

    I've tried several ways again. This last one I tried multiplying by the complex conjugate but that led me nowhere near the answer I needed unless I messed up (shown below). I'm not gonna type all of the work I did because it's just too tedious doing the latex. I basically got the it down to:

    \frac {sin}{cos} + \frac{1-sin}{cos}
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    \frac{sint}{cost} + \frac{cost}{1+sint} = \frac{sint(1+sint) + cost\cdot cost}{cost(1+sint)} =  \frac{sint + sin^2t + cos^2t}{cost(1+sint)} = \frac{sint + 1}{cost(sint+1)} = \frac{1}{cost} = sect
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    Quote Originally Posted by A Beautiful Mind View Post
    tan+\frac{cos}{1+sin} = sec

    I've tried several ways again. This last one I tried multiplying by the complex conjugate but that led me nowhere near the answer I needed unless I messed up (shown below). I'm not gonna type all of the work I did because it's just too tedious doing the latex. I basically got the it down to:

    \frac {sin}{cos} + \frac{1-sin}{cos}
    What do you mean by "complex conjugate" here? There are no complex numbers in this problem!

    I think what I would do is first change everything to sine and cosine (just because I am more comfortable with them than with the other functions!):
    \frac{sin(x)}{cos(x)}+ \frac{cos(x)}{1+ sin(x)}= \frac{1}{cos(x)}

    Now, get common denominators for the fractions on the left: \frac{sin(x)}{cos(x)}+ \frac{cos(x)}{1+ sin(x)}= \frac{sin(x)(1+ sin(x))}{cos(x)(1+ sin(x))}+ \frac{cos(x)(cos(x)}{cos(x)(1+ sin(x))} = \frac{sin(x)+ sin^2(x)}{cos(x)(1+ sin(x)}+ \frac{cos^2(x)}{cos(x)(1+ sin(x))}.

    Add the two fractions and use the fact that cos^2(x)+ sin^2(x)= 1 and you are home!
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