HELP NEEDED!!! WILL FAN AND MEDAL!!!! I need help understanding how to prove identities.

- anonymous

HELP NEEDED!!! WILL FAN AND MEDAL!!!! I need help understanding how to prove identities.

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- jamiebookeater

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- anonymous

I have a specific example. Please help...

- anonymous

what is it?

- anonymous

\[\tan \frac{ \cos \theta }{ 1+ \sin \theta }=\frac{ 1 }{ \cos \theta }\]

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## More answers

- anonymous

It was actually supposed to be tangent theta times cosine theta over one plus sine theta

- anonymous

I have all of the identities written out so I can easily refer to them, but the problem I am having is learning how to combine the like terms.

- anonymous

\[\frac{ \sin \theta }{ \cos \theta }+\frac{ \cos \theta }{ 1+ \sin \theta }\] this is what I need help combining because I have already converted Tan

- anonymous

what the heck is the tan doinog there ?

- anonymous

The tan was there because our teacher was trying to get us to recreate all of the identities to where they equal one another

- anonymous

no don't convert to tangent
in fact, if it was tangent, you would convert to \(\frac{\sin(x)}{\cos(x)}\)

- anonymous

lets do a tiny bit of algebra first

- anonymous

That's what I did and that was my farthest point in this.

- anonymous

ok i got that wrong, hold the phone

- anonymous

\[\frac{b}{a}+\frac{a}{1+b}\] that's better
can you add these?

- anonymous

I have no idea how to... or at least I can't remember. I know with multiplying you foil it, if you divide you use the kcf format :/ but adding and subtracting makes no sense to me because all I know is a common denominator is required

- anonymous

lol that is always the problem, the algebra

- anonymous

here is the one true way to add fractions \[\huge \frac{A}{B}+\frac{C}{D}=\frac{AD+BC}{BD}\]

- anonymous

o.o"

- anonymous

in your case you will have \[\frac{b(1+b)+a^2}{a(1+b)}\]

- anonymous

I am getting more lost by the minute...

- anonymous

that is the way to add fractions if they are numbers\[\frac{2}{7}+\frac{3}{5}=\frac{2\times 5+7\times 3}{7\times 5}\]

- anonymous

and it is also the way to add fractions if they have variables
it is the only way to do this, by using algebra to add

- anonymous

ohhhh ok, I get it, sort of.

- anonymous

forget that "least common denominator" nonsense you were taught
that is the way to add, you cannot avoid it

- anonymous

so ready to start again?

- anonymous

Yeah I suppose

- anonymous

using the one true way to add \[\frac{b}{a}+\frac{a}{1+b}\] you get \[\frac{b(1+b)+a^2}{a(1+b)}\]

- anonymous

now some more algebra, but just a little \[\frac{b(1+b)+a^2}{a(1+b)}=\frac{b+b^2+a^2}{a(1+b)}\]

- anonymous

that is just the distributive law in the numerator
now that we are done with algebra, we can go back to sines and cosines

- anonymous

sooooooo then \[\frac{ \cos \theta }{ \sin \theta } + \frac{ \cos \theta }{ 1+\sin \theta }\]
should become....

- anonymous

\[\frac{\sin(x)+\sin^2(x)+\cos^2(x)}{\cos(x)(1+\sin(x))}\]

- anonymous

Ok. But now comes an even trickier part.... I have to find a way to reduce that

- anonymous

hold a sec
the initial question was \[\frac{ \sin \theta }{ \cos \theta }+\frac{ \cos \theta }{ 1+ \sin \theta }\] right?

- anonymous

Yes

- anonymous

so after the algebra we get to \[\frac{\sin(x)+\sin^2(x)+\cos^2(x)}{\cos(x)(1+\sin(x))}\] don't look to reduce yet

- anonymous

the usual miracle occurs, \[\sin^2(x)+\cos^2(x)=1\] the mother of all trig identities

- anonymous

correct

- chrisplusian

The reasoning behind adding and subtracting the fractions is exactly what was said above, but the step being left out from above that might make it clear is that you just multiply both fractions by a number that will make both denominators the same or "common". In the above example....\[\frac{ b }{ a }+\frac{ a }{ 1+b } = (\frac{ 1+b }{ 1+b })(\frac{ b }{ a })+(\frac{ a }{ a })(\frac{ a }{ 1+b })\] and that is where the other equation @satellite73 came up with came from. Not to inturrupt

- anonymous

\[\frac{\sin(x)+\overbrace{\sin^2(x)+\cos^2(x)}^{\text{ this is 1}}}{\cos(x)(1+\sin(x))}\]

- anonymous

\[\frac{\sin(x)+1}{\cos(x)(1+\sin(x))}\] NOW you can cancel the common factor of \[1+\sin(x)\] top and bottom

- anonymous

Ok, ok, I see it now. so the Sins actually match and leave you with \[\frac{ 1 }{ \cos \theta }\]

- anonymous

yes
if you want to impress your teacher, write it as \[\sec(\theta)\]

- anonymous

i hope you also see that 98% of this is algebra

- anonymous

which, if your algebra is not what it needs to be, is going to be a problem, so bone up on it
also use the gimmick of replacing sine and cosine by letters to make the algebra easier
if we had to do this writing sine and cosine each time it would have been a pain

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