Question

1/1+cos(theta)+1/1-cos(theta)=2+2 cot^2(theta)

Answer 1

Seems doable. But @Rowel .. I need to know you're actually here, and that you're not a wall :D

Answer 2

\[\frac{ 1 }{ 1+\cos \theta }+\frac{ 1 }{ 1-\cos \theta }=2+2\cot ^{2}\theta \]

what is your objective here? We you asked to prove this?

Answer 3

I would believe so. It's an identity~

Answer 4

can you prove or verify 1/1+cos(theta)+1/1-cos(theta)=2+ cot^2 (theta)?

Answer 5

Of course I can. Now the task is to make sure you'd be able to do so too ^_^

Answer 6

can you help to verify?

Answer 7

But of course.
First, simplify the left-hand side.
You can do this by combining the sum of the two fractions into one fraction.

Normally, like this:
\[\Large \frac1a + \frac1b = \frac{a+b}{ab}\]

Answer 8

help me pls...show me the solution pls?

Answer 9

I *am* helping you.
But help me to help you ^_^

Come on, you can do this.. just simplify the left-side

Answer 10

i really dont know?pls....coz iam become crazy,,,,and i have 1 more question is 1/1+cos(theta)+1/1-cos(theta)=2+2 cot^2(theta)

Answer 11

veriy also csc^2(theta)/cot(theta)=csc(theta) sec (theta)

Answer 12

^ is squared rigth

Answer 13

Yes. But we really mustn't jump to the next question when you've barely started with the first :)

Answer 14

ok so help pls to verify my first question?pls?

Answer 15

Yes, I will help you, definitely, but you have to do your part too :)

Thus far, I have only asked you to simplify
\[\Large \frac1{\color{red}{1+\cos(\theta)}}+\frac1{\color{blue}{1-\cos(\theta)}}\]

Just like this:\[\Large \frac{1}{\color{red}a}+\frac{1}{\color{blue}b}= \frac{\color{red}a+\color{blue}b}{\color{red}a\color{blue}b}\]

Just compare and replace ^_^

Answer 16

ok tnx a lot?

Answer 17

You're welcome... though the question isn't answered yet. Can you take it from here?

Answer 18

yes i know?

Answer 19

csc²θ = csc θ sec θ
cot θ

Answer 20

verify pls?

Answer 21

This is actually simpler.
All you need are these simple facts:
\[\Large \csc(\theta) = \frac{1}{\sin(\theta)}\]\[\Large \sec(\theta) = \frac1{\cos(\theta)}\]

Answer 22

TJ, Rowel doesn't seem like he wants the help. To me it appears as if he wants you to tell him the answer. He hasn't realized that Maths doesn't work that way. He must work through it in order to understand it.

Answer 23

can you tell the rigth answer of what am i asking last night..?this is my assignment?tnx a lot?

Answer 24

i need help pls,,i am running out of time,,,,plsss...

Answer 25

Use i/a +i/b = a+b/ab to get \[\frac{ 1 }{ 1+\cos \theta } + \frac{ 1 }{ 1- \cos \theta} = \frac{ 2 }{ (1+\cos \theta ) (1-\cos \theta)}\]
\[\frac{ 1 }{ 1+\cos \theta } + \frac{ 1 }{ 1- \cos \theta} = \frac{ 2 }{ sin^2 \theta} = 2 csc^2 \theta\]
But \[\ csc^2 \theta = 1+ cot^2 \theta \]

So,
\[\frac{ 1 }{ 1+\cos \theta } + \frac{ 1 }{ 1- \cos \theta} = 2 + 2 cot^2 \theta\]

Answer 26

thank you so much,,,godbless you always my friend...actually..i hate math subject coz im become crazy..thanks for a big help...

Answer 27

how about this @shaikumar can ypu verify also this,,,,csc squared theta/cot theta=csc theta sec theta?