Prove the Identity (cosecA+cotA)^2 similar to 1+cosA/1-cosA

[ibysaiyan]

Homework Statement

Hi all , again i am stuck onto this question :( , tried over 3 sheets alone on it lol.btw. thanks for your replies ;) .
Prove the Identity (cosecA+cotA)^2 similar to 1+cosA/1-cosA

Homework Equations

hmm let's see.. sin2+cos2=1 ,
sec2= 1+tan2
cosec2= 1+cot2, cot=1/tanx= cosx/sinx

The Attempt at a Solution

i started off my expanding brackets:
=> cosec2A+cot2A+2cosecAcotA
and the rest well =/ no idea.. i couldn't prove them, just had a thought while typing can't i like sub . some value before expanding them ? ooh =/

 

[rock.freak667]

Write everything in terms of sinA and cosA, then simplify some more.

 

[Mark44]

Homework Statement

Hi all , again i am stuck onto this question :( , tried over 3 sheets alone on it lol.btw. thanks for your replies ;) .
Prove the Identity (cosecA+cotA)^2 similar to 1+cosA/1-cosA

Do you mean "=" where you have written "similar to?"
Also, you need parentheses on the right side. What you have would be correctly interpreted as 1 + (cosA/1) - cosA, but I'm pretty sure that's not what you intended.

Homework Equations

hmm let's see.. sin2+cos2=1 ,
sec2= 1+tan2
cosec2= 1+cot2, cot=1/tanx= cosx/sinx

At least give some hint that you are talking about the squares of functions. Without any context sin2 would be interpreted as the sine of 2 radians.

The Attempt at a Solution

i started off my expanding brackets:
=> cosec2A+cot2A+2cosecAcotA

You started with (cscA + cotA)^2. This is equal to csc^2(A) + 2cscAcotA + cot^2(A). When you write cosec2A, most people would read this as cosec(2A) (usually written as csc(2A)).

and the rest well =/ no idea.. i couldn't prove them, just had a thought while typing can't i like sub . some value before expanding them ? ooh =/

 

[ibysaiyan]

Ah. again thanks a zillion :)
i think i got it..:
(cosecA + cotA) ^2
(1/SinA + CosA/SinA )^2
=>(1+cosA^2 /sin^2A) <-- i am doubtful about this bit
=> 1+cos2A/1-cos2A

 

[ibysaiyan]

oh sorry about that, next time i will use latex.

 

[Mark44]

Ah. again thanks a zillion :)
i think i got it..:
(cosecA + cotA) ^2
(1/SinA + CosA/SinA )^2
=>(1+cosA^2 /sin^2A) <-- i am doubtful about this bit
=> 1+cos2A/1-cos2A

Here's what you want to say
(cosecA + cotA) ^2 = (1/SinA + CosA/SinA )^2
= [(1 + cosA)/sinA]^2
When you square 1 + cosA you will have three terms. What you have below has only two terms.

= (1+cosA^2 /sin^2A) <-- i am doubtful about this bit

 

[ibysaiyan]

Here's what you want to say
(cosecA + cotA) ^2 = (1/SinA + CosA/SinA )^2
= [(1 + cosA)/sinA]^2
When you square 1 + cosA you will have three terms. What you have below has only two terms.

yea (a+b)^2 formula so am i on the right track so far?

 

[Mark44]

Up to that point, yes.

 

[ibysaiyan]

yea (a+b)^2 formula so am i on the right track so far?

this is what i got:
[(1+cosA / Sin A )]^2
=> 1+ cos^2A + 2cosA / Sin^2 A
=> hmm would i take cos as common?

 

[Mark44]

this is what i got:
[(1+cosA / Sin A )]^2
=> 1+ cos^2A + 2cosA / Sin^2 A
=> hmm would i take cos as common?

You are using ==> ("implies") when you should be using =.

You have
\frac{(1 + cosA)^2}{sin^2A}
Leave the numerator in factored form (don't expand it).
Rewrite the denominator using an identity that you know.
Factor the denominator.
Simplify.

 

[ibysaiyan]

AH... thanks a lot , i get it now( much clearer ).
Solved .

 

[Mark44]

Your work that you turn in should look pretty much like this:
(cscA + cotA)^2~=~(\frac{1}{sinA}~+~\frac{cosA}{sinA})^2~=~\frac{(1 + cosA)^2}{sin^2A}~=~\frac{(1 + cosA)^2}{(1 - cosA)^2}~=~\frac{(1 + cosA)(1 + cosA)}{(1 - cosA)(1 + cosA)}~=~\frac{1 + cosA}{1 - cosA}

To see my LaTeX code, click what I have above and a new window will open that has my script.

 

[ibysaiyan]

Your work that you turn in should look pretty much like this:
(cscA + cotA)^2~=~(\frac{1}{sinA}~+~\frac{cosA}{sinA})^2~=~\frac{(1 + cosA)^2}{sin^2A}~=~\frac{(1 + cosA)^2}{(1 - cosA)^2}~=~\frac{(1 + cosA)(1 + cosA)}{(1 - cosA)(1 + cosA)}~=~\frac{1 + cosA}{1 - cosA}

To see my LaTeX code, click what I have above and a new window will open that has my script.

Roger that , to be honest i can't thank you enough =), anyhow i will be back again ( that's for sure using LATEX code) =).