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# How is this equation an identity?

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Here's the equation: $${sin(x)cos(x) \over 1 + cos(x)}$$=$${1-cos(x) \over tan(x)}$$

So the non-permissible values are x ≠ $${π \over 2}$$ + πn, n ∈ I; x ≠ π + 2πn, n ∈ I

they are very distinct.

I graphed both of these expressions on desmos.com and I'm just very confused as to how these graphs are identical? If the non-permissible values for the first and second expression are different than how are the graphs the same when graphed?

Thank you! :)

Feb 19, 2019
edited by Guest  Feb 19, 2019

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I am not sure what you problems is. I have graphed them here

https://www.desmos.com/calculator/nhxdil3szg

As you can see, the two graphs are identical.  (except for the holes)

Did you want to see how they are the same algebraically?

Oh, you are saying that since the non permissable values are different the graphs cannot be the same.

You are right.  The identity should state that it is not true for those points.

Feb 19, 2019
edited by Melody  Feb 19, 2019
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So are the holes and the vertical dotted lines the non-permissible values?

Guest Feb 20, 2019
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Umm

The asyptotes are true for both graphs so they do not effect the identity.

The holes are only present for the right hand side so they are the ones that should be stated as non-permissable for the identity.

Melody  Feb 20, 2019
edited by Melody  Feb 20, 2019