My Math Forum Reconcile 2 Trigonometric Indentities

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 June 27th, 2019, 09:43 AM #1 Newbie   Joined: Jan 2017 From: London Posts: 10 Thanks: 0 Reconcile 2 Trigonometric Indentities Hi All, I'm stuck on a particular problem set: Establish the validity (reconcile) of the following - (1+sin x)/cos x = cos x/(1-sin x) I'm finding it difficult to walk from either left to right or right to left i.e. I cannot decompose into other identities that will lead me to the other side. I thought about expanding the left side to sec x + tan x but that leads me nowhere. I'm stuck!! Help!! Thanks Last edited by skipjack; June 28th, 2019 at 11:05 PM.
 June 27th, 2019, 09:53 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 just get both sides over a common denominator and it should be obvious. Thanks from topsquark and mathsonlooker
June 27th, 2019, 10:12 AM   #3
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 Originally Posted by romsek just get both sides over a common denominator and it should be obvious.
I think cross-multiply to eliminate the fractions altogether gets them there even faster, no?

June 27th, 2019, 10:23 AM   #4
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 Originally Posted by DarnItJimImAnEngineer I think cross-multiply to eliminate the fractions altogether gets them there even faster, no?
they are identical operations

June 28th, 2019, 08:37 AM   #5
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 Originally Posted by romsek they are identical operations
They are, but one I can do easily in my head, and with the other I run out of memory.

 June 28th, 2019, 08:43 AM #6 Newbie   Joined: Jan 2017 From: London Posts: 10 Thanks: 0 Thanks @romsek @DarnItJimImAnEngineer. Making the denominator common amongst both and crossing out common terms does make it obvious. I also took another path; I split (1-sin x) into (1-sin^2x)/(1+sin x) on the right hand side. Then cos x/[(1-sin^2x)/(1+sin x)] (using the identity cos^2x+ sin^2x =1) and one can simplify this to what's on the left = (1+sin x)/cos x Last edited by skipjack; June 28th, 2019 at 10:33 AM.
 June 28th, 2019, 11:09 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2205 Does that always work? Thanks from topsquark
 June 28th, 2019, 06:51 PM #8 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond $\displaystyle \frac{1+\sin x}{\cos x}\cdot\frac{1-\sin x}{1-\sin x}=\frac{1-\sin^2x}{\cos x(1-\sin x)}=\frac{\cos^2x}{\cos x(1-\sin x)}=\frac{\cos x}{1-\sin x}$ Thanks from mathsonlooker
 June 28th, 2019, 11:06 PM #9 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2205 What exactly does that prove? Last edited by skipjack; June 28th, 2019 at 11:30 PM.
June 30th, 2019, 02:31 PM   #10
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 Originally Posted by greg1313 $\displaystyle \frac{1+\sin x}{\cos x}\cdot\frac{1-\sin x}{1-\sin x}=\frac{1-\sin^2x}{\cos x(1-\sin x)}=\frac{\cos^2x}{\cos x(1-\sin x)}=\frac{\cos x}{1-\sin x}$
Interesting!

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