Trig

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June 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!!!
June 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
June 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?
June 10th 2012, 03:10 AM
srirahulan

Re: Trig

pi is a constant.i am really sorry to undefind it.
June 10th 2012, 03:46 AM
BobP

Re: Trig

The question doesn't seem to make much sense anyway, doesn't $\frac{\cos(A+\pi)}{\cos(A-\pi)}=1$ ?
June 10th 2012, 06:47 AM
Tclack

1 Attachment(s)

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

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