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 July 12th, 2011, 10:47 PM #1 Newbie   Joined: Jul 2011 Posts: 7 Thanks: 0 Trig Identities Here's another one if you're up for it mark tan x 1 ------ = --------- sin x - 2tan x cos x - 2
 July 12th, 2011, 10:48 PM #2 Newbie   Joined: Jul 2011 Posts: 7 Thanks: 0 Re: Trig Identities darn that didn't work out (tan x)/(sin x - 2tan x) = 1/(cos x - 2)
 July 12th, 2011, 10:49 PM #3 Newbie   Joined: Jul 2011 Posts: 7 Thanks: 0 Re: Trig Identities *prove that gah sorry for the triple post I cant edit my own posts
 July 12th, 2011, 10:58 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: Trig Identities We are given: $\frac{\tan x}{\sin x-2\tan x}=\frac{1}{\cos x-2}$ On the left side, try multiplying by $1=\frac{\cot x}{\cot x}$
July 12th, 2011, 11:02 PM   #5
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Re: Trig Identities

Quote:
 Originally Posted by Gokias ...I cant edit my own posts
Once you have made 10 or more posts you will be able to. It is an unfortunate but necessary anti-spam measure taken by our admin. I apologize for the temporary inconvenience.

July 13th, 2011, 08:06 AM   #6
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Re: Trig Identities

Hello, Gokias!

Another approach . . .

Quote:
 $\text{Prove: }\:\frac{\tan x}{\sin x\,-\,2\tan x} \:=\:\frac{1}{\cos x\,-\,2}$

$\text{W\!e have: }\:\frac{\tan x}{\sin x\,-\,2\tan x} \;=\;\frac{\frac{\sin x}{\cos x}}{\sin x\,-\,2\,\frac{\sin x}{\cos x}}$

$\text{Multiply by }\frac{\cos x}{\cos x}:\;\;\frac{\cos x\left(\frac{\sin x}{\cos x}\right)}{\cos x\left(\sin x\,-\,2\,\frac{\sin x}{\cos x}\right)} \;=\;\frac{\sin x}{\sin x\cos x\,-\,2\sin x} \;=\;\frac{\cancel{\sin x}}{\cancel{\sin x}(\cos x\,-\,2)} \;=\;\frac{1}{\cos x\,-\,2}$

 July 13th, 2011, 08:37 AM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,900 Thanks: 1093 Math Focus: Elementary mathematics and beyond Re: Trig Identities $\frac{\tan(x)}{\sin(x)\,-\,2\tan(x)}\,\cdot\,\frac{cos(x)\,-\,2}{\cos(x)\,-\,2}\,=\,\frac{1}{cos(x)\,-\,2}$
July 13th, 2011, 09:07 AM   #8
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Re: Trig Identities

Quote:
 Originally Posted by greg1313 $\frac{\tan(x)}{\sin(x)\,-\,2\tan(x)}\,\cdot\,\frac{cos(x)\,-\,2}{\cos(x)\,-\,2}\,=\,\frac{1}{cos(x)\,-\,2}$
As long as cos(x) ? 2

 July 13th, 2011, 09:24 AM #9 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,900 Thanks: 1093 Math Focus: Elementary mathematics and beyond Re: Trig Identities
 July 13th, 2011, 01:20 PM #10 Senior Member   Joined: Sep 2010 Posts: 101 Thanks: 0 Re: Trig Identities Hahahaha, good one!

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