# Trig

• Jun 9th 2012, 07:29 PM
srirahulan
Trig
If $tan A=\ sec2\alpha\ cot\pi$ prove that $\frac{cos(A+\pi)}{cos(A-\pi)}=-tan^2 \alpha$ I tried this by simplifying the statement and finally i get $\frac{cos(A - \pi)}{cosAcos\pi}=\frac{2cos^2\alpha}{2cos^2\alpha-1}$ then i can't simplify this prove.please help me!!!
• Jun 9th 2012, 07:33 PM
Amer
Re: Trig
do you mean by $\pi$ the known pi or it is just another constant if it is the known pi then

$\cot \pi$ is undefind
• Jun 9th 2012, 07:53 PM
Tclack
Re: Trig
Quote:

Originally Posted by srirahulan
If $tan A=\ sec2\alpha\ cot\pi$

is that secant supposed to be squared?
• Jun 10th 2012, 03:10 AM
srirahulan
Re: Trig
pi is a constant.i am really sorry to undefind it.
• Jun 10th 2012, 03:46 AM
BobP
Re: Trig
The question doesn't seem to make much sense anyway, doesn't $\frac{\cos(A+\pi)}{\cos(A-\pi)}=1$ ?
• Jun 10th 2012, 06:47 AM
Tclack
Re: Trig
Well, IF you meant sec squared(you didn't get back to me, so idk), then maybe This looks like it's getting somewhere

Attachment 24057
it looks like the right side will reduce to -1, so you can substitute in that cos(A+pi)/cos(A-pi) = 1 from above