$\displaystyle (1 + csc(-x))/(1 + sin(-x)) = sec(x) + tan(x)$

what do i do here? I keep forgetting my trig identities :/

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- Nov 17th 2009, 04:24 PM>_<SHY_GUY>_<trig identities II
$\displaystyle (1 + csc(-x))/(1 + sin(-x)) = sec(x) + tan(x)$

what do i do here? I keep forgetting my trig identities :/ - Nov 17th 2009, 04:38 PMskeeter
- Nov 17th 2009, 04:45 PM>_<SHY_GUY>_<
- Nov 17th 2009, 04:52 PMskeeter
one more time, the equation you posted in

__not__an identity. in mathematical terms ...

$\displaystyle \frac{1 + \csc(-x)}{1 + \sin(-x)} \ne \sec{x} + \tan{x}$

clear? - Nov 17th 2009, 04:54 PMANDS!
On the right side:

$\displaystyle sec(x)+tan(x)$

$\displaystyle \frac{1}{cos(x)}+\frac{sin(x)}{cos(x)}$

$\displaystyle \frac{cos(x)+cos(x)sin(x)}{cos^{2}(x)}$

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

$\displaystyle \frac{cos(x)(1+sin(x)}{(1+sin(x))(1-sin(x))}$

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

$\displaystyle \frac{cos(x)}{1+sin(-x)}$

I'm a bit rusty on my trig as well, but maybe the above can help you. Got that errant cosine that you somehow need to manipulate. Work from the left hand side, and see if you get to any of the steps I did on the right hand side. - Nov 17th 2009, 04:57 PM>_<SHY_GUY>_<
the book asks to verify the identity, so could it be an error in the book?

- Nov 17th 2009, 04:58 PMANDS!
Is the problem written correctly? I'm willing to bet there is no error in the book you are using (or rather there might be, but the likelihood that we're working out a problem that is isn't too high).

- Nov 17th 2009, 05:09 PM>_<SHY_GUY>_<