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-   -   Reconcile 2 Trigonometric Indentities (http://mymathforum.com/trigonometry/346662-reconcile-2-trigonometric-indentities.html)

 mathsonlooker June 27th, 2019 09:43 AM

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

 romsek June 27th, 2019 09:53 AM

just get both sides over a common denominator and it should be obvious.

 DarnItJimImAnEngineer June 27th, 2019 10:12 AM

Quote:
 Originally Posted by romsek (Post 611133) 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?

 romsek June 27th, 2019 10:23 AM

Quote:
 Originally Posted by DarnItJimImAnEngineer (Post 611135) I think cross-multiply to eliminate the fractions altogether gets them there even faster, no?
they are identical operations

 DarnItJimImAnEngineer June 28th, 2019 08:37 AM

Quote:
 Originally Posted by romsek (Post 611136) they are identical operations
They are, but one I can do easily in my head, and with the other I run out of memory.

 mathsonlooker June 28th, 2019 08:43 AM

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

 skipjack June 28th, 2019 11:09 AM

Does that always work?

 greg1313 June 28th, 2019 06:51 PM

$\displaystyle \frac{1+\sin x}{\cos x}\cdot\frac{1-\sin x}{1-\sin x}=\frac{1-\sin^2x}{\cos x(1-\sin x)}=\frac{\cos^2x}{\cos x(1-\sin x)}=\frac{\cos x}{1-\sin x}$

 skipjack June 28th, 2019 11:06 PM

What exactly does that prove?

 mathsonlooker June 30th, 2019 02:31 PM

Quote:
 Originally Posted by greg1313 (Post 611194) $\displaystyle \frac{1+\sin x}{\cos x}\cdot\frac{1-\sin x}{1-\sin x}=\frac{1-\sin^2x}{\cos x(1-\sin x)}=\frac{\cos^2x}{\cos x(1-\sin x)}=\frac{\cos x}{1-\sin x}$
Interesting!

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