how would i begin to prove the equation 2/1+cosx - tan^2 x/2=1

- anonymous

how would i begin to prove the equation 2/1+cosx - tan^2 x/2=1

- schrodinger

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- xapproachesinfinity

so it is \[\frac{1}{1+\cot x}-\tan^2x=1\]

- anonymous

okay i understand

- xapproachesinfinity

the first thing im thinking about is multiplying by 1-cotx top and bottom for the 1/1+cotx

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## More answers

- xapproachesinfinity

i haven't started yet i was just asking is that the right thing?
because the way you wrote it can be misleading

- xapproachesinfinity

oh hold on i missed 2 on top

- anonymous

well it's like \[(2/1+cosx) - \tan^2 x/2=1\]

- xapproachesinfinity

\[\frac{2}{1+\cot x}-\frac{\tan^2x}{2}=1\]

- xapproachesinfinity

may bad! cos not cot

- anonymous

right right i got it

- xapproachesinfinity

i must be blind or something lol

- xapproachesinfinity

\[\frac{2}{1+\cos x}-\frac{\tan^2x}{2}=1\]
looking at the left side
\[\frac{2}{1+\cos x}-\frac{\tan^2x}{2}=\frac{2(1-\cos x)}{(1+\cos x)(1-\cos x)}-\frac{\tan^2x}{2}\]

- xapproachesinfinity

let's go from there

- xapproachesinfinity

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

- xapproachesinfinity

can you go further your self

- xapproachesinfinity

actually that's not an identity

- anonymous

i didnt think so lol

- xapproachesinfinity

oh no i take it back it is!
when i plugged in 0
i forgot something lol

- xapproachesinfinity

hmm for pi/4 it is not good!

- xapproachesinfinity

are you sure you typed the right thing!

- anonymous

i will show you exactly what's in the textbook but how do you get the horizontal division line in the equation like you did in all your formulas?

- xapproachesinfinity

use drawing tool if you have a mouse you will do find drawing the problem

- xapproachesinfinity

snap shot your problem don't you have a phone?

- xapproachesinfinity

what you have there is not an identity
i verified it with couple values and it fails miserably
if it is an identity it should not feel even one

- xapproachesinfinity

fail*

- anonymous

\[\frac{ 2 }{ 1 + cosx } - \tan^2\frac{ x }{ 2 } = 1\]

- xapproachesinfinity

ah wow why you didn't correct me above
i was dividing tan^2 over 2
2 was supposed to be within tan

- xapproachesinfinity

darn you made me work really hard lol

- anonymous

im sorry i wouldve if i known it was wrong but i was confused on this entire problem hence why i am here asking for help

- xapproachesinfinity

here what you can do
\[\frac{2}{1+\cos x}=1+\tan^2(x/2)=\sec^2(x/2)\]
if you prove the left is sec^2(x/2) then you are done!

- xapproachesinfinity

that is the same as proving
\[\frac{1+\cos x}{2}=\cos^2(x/2) \]

- xapproachesinfinity

isn't that some familiar identity to yoou?

- xapproachesinfinity

the last line a i wrote should be obvious to you

- anonymous

i get it i think the sec and tan identity was what i was thinking i would get to to prove the identity i just wasnt sure how to get there

- xapproachesinfinity

1+cosx/2=cos^2(x/2)
is called half angle identity
which can be proven easily
but you don't need to draw it all the way
as long as that is an identity we know of
so the initial problem is an identity as well

- anonymous

right right i remember that now

- xapproachesinfinity

let's prove that half angle identity for the sake of argument
we know that cosx=2cos^2(x/2)-1
so 1+cosx=2cos^2(x/2)
cos^2(x/2)=(1+cosx)/2 hence proved
[cos(2a)=2cos^2a-1 is another useful identity]

- xapproachesinfinity

they are all linked!

- xapproachesinfinity

oh well gotta leave!

- anonymous

how do i prove it's sec though?

- anonymous

@zepdrix I was wondering if you can demonstrate this identity
2/1+cosx - tan^2 x/2=1

- anonymous

I'm pretty new to this myself, so I'm trying to understand what @xapproachesinfinity did
but I think I'm kinda lost...

- zepdrix

Mmm can we do it in a new thread? :D
I don't feel like reading through all of these comments XD lol

- anonymous

mm I pretty much a noob in trig identity LOL
but ok xD

- xapproachesinfinity

@Learner1298 we already proved it?
why you need to prove sec lol
read through what i did

- Loser66

\[\dfrac{2}{1+cos (x) }-tan^2(x/2)\] \[= \dfrac{2}{1+cos(x)} -\dfrac{sin^2(x)}{(1+cos (x))^2}\]
because \(tan (x/2) = \dfrac{sin (x) }{1+cos x}\)

- Loser66

hence, it turns to
\(\dfrac{2(1+cos(x)-sin^2(x)}{(1+cos(x))^2}\)
Now, numerator only:

- Loser66

\(2(1+ cos(x) ) -sin^2(x)\\=2+2cosx-(1-cos^2(x) =2+2cos(x)-1+cos^2(x) \\=1+2cos(x) +cos^2(x) = (1+cos(x))^2 \)

- Loser66

Put it back to the fraction above, we can see that numerator = denorminator , hence the fraction =1 = R H S

- Loser66

@xapproach.... hey kid, the second reader didn't understand your stuff. Let old man helps you out. hahahaha.....................

- xapproachesinfinity

you did a different version lol
this trig stuff there are often many diff ways of doing stuff

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