# Thread: verifying identities...another one i'm stuck on.

1. ## 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:

$\frac {sin}{cos}$ $+$ $\frac{1-sin}{cos}$

2. $\frac{sint}{cost} + \frac{cost}{1+sint} = \frac{sint(1+sint) + cost\cdot cost}{cost(1+sint)} =$ $\frac{sint + sin^2t + cos^2t}{cost(1+sint)} = \frac{sint + 1}{cost(sint+1)} = \frac{1}{cost} = sect$

3. Originally Posted by A Beautiful Mind
$tan+\frac{cos}{1+sin} = sec$

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!