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June 27th, 2019, 09:43 AM   #1
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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!!

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Last edited by skipjack; June 28th, 2019 at 11:05 PM.
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June 27th, 2019, 09:53 AM   #2
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just get both sides over a common denominator and it should be obvious.
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June 27th, 2019, 10:12 AM   #3
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Quote:
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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?
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June 27th, 2019, 10:23 AM   #4
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I think cross-multiply to eliminate the fractions altogether gets them there even faster, no?
they are identical operations
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June 28th, 2019, 08:37 AM   #5
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they are identical operations
They are, but one I can do easily in my head, and with the other I run out of memory.
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June 28th, 2019, 08:43 AM   #6
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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.
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June 28th, 2019, 11:09 AM   #7
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Does that always work?
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June 28th, 2019, 06:51 PM   #8
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$\displaystyle \frac{1+\sin x}{\cos x}\cdot\frac{1-\sin x}{1-\sin x}=\frac{1-\sin^2x}{\cos x(1-\sin x)}=\frac{\cos^2x}{\cos x(1-\sin x)}=\frac{\cos x}{1-\sin x}$
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June 28th, 2019, 11:06 PM   #9
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What exactly does that prove?

Last edited by skipjack; June 28th, 2019 at 11:30 PM.
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June 30th, 2019, 02:31 PM   #10
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Originally Posted by greg1313 View Post
$\displaystyle \frac{1+\sin x}{\cos x}\cdot\frac{1-\sin x}{1-\sin x}=\frac{1-\sin^2x}{\cos x(1-\sin x)}=\frac{\cos^2x}{\cos x(1-\sin x)}=\frac{\cos x}{1-\sin x}$
Interesting!
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