can anyone help with this problem?

Prove

tan^2 + cos^2 + sin^2 = 1/cos^2

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- Jan 10th 2007, 04:51 PMimthatgirlTrig identities
can anyone help with this problem?

Prove

tan^2 + cos^2 + sin^2 = 1/cos^2 - Jan 10th 2007, 05:34 PMThePerfectHacker
- Jan 10th 2007, 05:50 PMThePerfectHacker
$\displaystyle \tan^2 x - \cos^2 x= \frac{1}{\cos ^2 x}- 1 -\cos ^2 x$

$\displaystyle (\sec^2 x-1)-\cos^2 x=\frac{1}{\cos^2 x}-1-\cos^2 x$

$\displaystyle \sec^2 x - 1 - \cos^2 x= \frac{1}{\cos^2 x}-1 -\cos^2 x$

$\displaystyle \frac{1}{\cos^2 x}-1-\cos^2 x = \frac{1}{\cos^2 x}-1 -\cos^2 x$ - Jan 10th 2007, 06:21 PMimthatgirl
thank you so much for the help i get it now

- Jan 10th 2007, 10:30 PMearboth
Hello,

you have to use these properties:

$\displaystyle \cos^2(x)+\sin^2(x)=1$

$\displaystyle \tan(x)=\frac{\sin(x)}{\cos(x)}$

$\displaystyle \tan^2(x)+\cos^2(x)+\sin^2(x)=\tan^2(x)+1-\sin^2(x)+\sin^2(x)=$$\displaystyle

\tan^2(x)+1=\frac{\sin^2(x)}{\cos^2(x)}+1=\frac{\s in^2(x)+\cos^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}$

EB

PS: If you have a new problem please start a new thread. Otherwise you risk that nobody will notice that you asked for some help. - Jan 11th 2007, 03:56 AMtopsquark
I'll reinforce that comment. If you remove your original question then no one can see what you originally posted and thus be confused about what someone's response to it is. If someone else is looking for help on a similar topic, that makes it harder for them to learn it themselves. So please, when you have a new question either post a completely new thread, or just respond to the post you already started.

-Dan