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

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

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

ibysaiyan said:

## 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.
ibysaiyan said:

## 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.
ibysaiyan said:

## 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)).
ibysaiyan said:
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 =/

Ah. again thanks a zillion :)
i think i got it..:
(cosecA + cotA) ^2
(1/SinA + CosA/SinA )^2
=> 1+cos2A/1-cos2A

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

ibysaiyan said:
Ah. again thanks a zillion :)
i think i got it..:
(cosecA + cotA) ^2
(1/SinA + CosA/SinA )^2
=> 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.
ibysaiyan said:

Mark44 said:
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?

Up to that point, yes.

ibysaiyan said:
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?

ibysaiyan said:
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.

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

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.

Mark44 said:
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) =).

## 1. How do you prove the identity (cosecA+cotA)^2 similar to 1+cosA/1-cosA?

To prove this identity, we can start by expanding the left side of the equation using the trigonometric identities cosecA = 1/sinA and cotA = cosA/sinA. This will give us (1/sinA + cosA/sinA)^2, which can be simplified to (1+cosA)^2/sin^2A. Next, we can use the Pythagorean identity sin^2A + cos^2A = 1 to replace sin^2A in the denominator, giving us (1+cosA)^2/(1-cos^2A). Finally, we can use the identity 1-cos^2A = sin^2A to simplify further, resulting in (1+cosA)^2/(1-cosA)(1+cosA), which simplifies to 1+cosA/1-cosA, proving the identity.

## 2. What are the steps involved in proving this identity?

The steps involved in proving this identity are: 1) Expand the left side of the equation using trigonometric identities, 2) Simplify the expression using known identities, 3) Use the Pythagorean identity to replace sin^2A in the denominator, and 4) Simplify the expression further to match the right side of the equation.

## 3. Can you provide an example of how this identity is used in a mathematical problem?

Sure, for example, if we have an equation that involves the expression (cosecA+cotA)^2, we can use the identity to simplify it to 1+cosA/1-cosA, making it easier to solve or manipulate in the problem.

## 4. What is the significance of this identity in the field of trigonometry?

This identity is significant in the field of trigonometry because it allows us to simplify expressions and equations involving trigonometric functions, making them easier to work with and manipulate. It also helps us establish connections between different trigonometric identities, leading to a better understanding of the subject.

## 5. Are there any special cases or exceptions to this identity?

No, this identity holds true for all values of A, as long as the trigonometric functions involved are defined. However, it is important to note that when cosA = 1, the right side of the equation becomes undefined, so this identity cannot be applied in that case.

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