Thread: trig identities II

1. trig identities II

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

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

2. Originally Posted by >_<SHY_GUY>_<
$(1 + csc(-x))/(1 + sin(-x)) = sec(x) + tan(x)$

what do i do here? I keep forgetting my trig identities :/
note that this equation is not an identity.

$\frac{1+csc(-x)}{1+\sin(-x)} =$

$\frac{1-csc{x}}{1-\sin{x}} =$

$\frac{\sin{x} - 1}{\sin{x}(1-\sin{x})} =$

$-\frac{1}{\sin{x}} = -\csc{x}$

3. Originally Posted by skeeter
note that this equation is not an identity.

$\frac{1+csc(-x)}{1+\sin(-x)} =$

$\frac{1-csc{x}}{1-\sin{x}} =$

$\frac{\sin{x} - 1}{\sin{x}(1-\sin{x})} =$

$-\frac{1}{\sin{x}} = -\csc{x}$
then how would you get sec(x) + tan(x) ? im sorry i just dont see what you did that has to do with the problem :/

4. one more time, the equation you posted in not an identity. in mathematical terms ...

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

clear?

5. On the right side:

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

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.

6. the book asks to verify the identity, so could it be an error in the book?

7. 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).

8. Originally Posted by ANDS!
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).

I believe its an error in the book because its written like that. Thank you for your help and I'll let you go because I feel like I'm bothering you too much. Thank you again(: