Question

I need help on simplifying a fraction.
TRIG.
I need the fastest way to do this.

I am usually able to do them, but this one I got stock on.

Answer 1

\(\Huge\color{blue}{ \sf \frac{Csc(x)+Cot(x) }{ Csc(x)-Cot(x)} }\) the first thing I did, is that I put everything in terms of sines and cosines, add the fractions that I got on top and bottom, and when I divided them I canceled Sin(x) on top and bottom. I got (now, as my result)

\(\Huge\color{blue}{ \sf \frac{1+Cos(x) }{ 1-Cos(x)} }\)

and then I broke my head (dky)

Answer 2

Try multiplying both numerator and denominator by (1+cos x) and see where that gets you.

Answer 3

Yeah, I tried to do that (ik this is conjugate)

\(\Huge\color{blue}{ \bf \frac{Cos^2(x)+2Cos(x)+1}{Sin^2(x)} }\)

Answer 4

Well, or 1-Cos^2x

Answer 5

I got disconnected.

Answer 6

(1+cos(x))/(1-cos(x))
then,
you can use these identities :
1+cosx= 2cos^2 (x/2) and
1-cosx= 2sin^2 (x/2)

Answer 7

I got disconnected, but I think I know how to do it...

Answer 8

Try leaving the numerator as (1+cos x)^2. This may or may not help.
There could be multiple ways of presenting your final result. Choose the one that you think is simplest.

Answer 9

Oh, logic, why must you be ambiguous? :D

What exactly are we supposed to be doing here?
How simple must simple be?

Like just one trigonometric function?

Answer 10

\(\Large\color{blue}{ \bf \frac{1+cos(x)}{1-cos(x)} }\) first I am going to use conjungate, (denominator is Sin^2(x) )
and multiply everything times Csc^2(x) saying that I will have nothing on the bottom


\(\large\color{blue}{ \bf Csc^2(x)~~(~Cos^2(x)+2Cos(x)+1~)}\) some monipulation.

\(\large\color{blue}{ \bf Csc^2(x)~~(~Cos^2(x)+2Cos(x)+Sin^2(x)+Cos^2(x)~~~)}\)

\(\large\color{blue}{ \bf Cot^2(x)+Cot^2(x)+ 2Cos(x)Csc^2(x)+1}\)

\(\large\color{blue}{ \bf 2Cot^2(x)+ 2Cot(x)Csc(x)+1}\)

idk, it said as simple as possible

Answer 11

We might have very different ideas on what constitute simple ;)
Although, I'd rather work exclusively with sines and cosines, for reasons I'm not really sure of :D

Answer 12

Okay, here's the deal...
\[\Large \cot\left(\frac x 2\right)\]
while looking particularly nasty... is going to be your best friend here :)

Answer 13

half angle? I really haven't thought about that -:(

Answer 14

it should be cot^2 (x/2)

Answer 15

\(\large\color{blue}{ \bf 2Cot^2(x)+ 2Cot(x)Csc(x)+1}\)

using 1+cot^2(x)=csc^2(x)

\(\large\color{blue}{ \bf Cot^2(x)+Csc^2(x)+ 2Cot(x)Csc(x)}\)

This is what my next step would have been.

yeah, cot^2 , right.

Answer 16

\[\Large \cot \frac x 2 = \frac{\cos \frac x2}{\sin\frac x2}= \frac{\sqrt{\frac{1-\cos(x)}{2}}}{\sqrt{\frac{1+\cos(x)}{2}}}\]

And d****t, @RadEn a spoiler alert next time? XD

Answer 17

I understand you are truly the best

Answer 18

Wait a minute... mixed up, hang on
\[\LARGE\cot \frac x 2 = \frac{\cos \frac x2}{\sin\frac x2}= \frac{\sqrt{\frac{1+\cos(x)}{2}}}{\sqrt{\frac{1-\cos(x)}{2}}}\]
that's better...

Answer 19

lol Zelman, appreciate the kudos, but I didn't mean it like that? ^_^
Peace :)

Answer 20

I got the idea, thank you!

Answer 21

i just remember 2 identities below
cos^2 (a) = (1+cos(2a)/2 and
cos^2 (a) = (1+cos(2a)/2

actually, by manipulation we can take that a = x/2

Answer 22

oppsss. mistake in vaste :)
cos^2 (a) = (1+cos(2a)/2
and
sin^2 (a) = (1-cos(2a)/2

Answer 23

Yeah, I get it.... I like how terenzreignz did it. It was just awesome :)))

Answer 24

...yeah... I suppose. :>
I have a name, it's TJ (not really), so use it next time, ok, Zelman? ^^

Peace ^_^

Answer 25

Yeah, terenzreignz :) Thank you for your help !

Answer 26

^mother of irony XD

Answer 27

Wait, what ?

Answer 28

I meant don't call me terenzreignz, actually :)

No real biggie, more like a pet peeve ^_^

Answer 29

Well, you called yourself that, but whatever you want, pet peeve

Answer 30

<sigh>
XD

From now on, just stick to "Terenz"

I just logged in to OS to help in this question... so, signing off now :D
---------
Terenz out

Answer 31

K