I am trying to learn how to transform identities of one side of the equation to the other side. is this correct?

I am using 0 for theta

theta = 0

question:

tan 0 + cot 0

-------------- = csc^2 0

tan 0

answer:

cot 0 = csc 0

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- Sep 14th 2008, 09:43 AMrobasctrigonometric identities
I am trying to learn how to transform identities of one side of the equation to the other side. is this correct?

I am using 0 for theta

theta = 0

**question:**

tan 0 + cot 0

-------------- = csc^2 0

tan 0

**answer:**

cot 0 = csc 0

- Sep 14th 2008, 09:50 AMJhevon
that answer is wrong. how does that prove an identity. you must show the left hand side and right hand side are the same. so you must turn one side into the other or bring both sides to the same thing

consider the left hand side (it is more complicated and so gives us more options to change things)

note, $\displaystyle \frac {\tan \theta + \cot \theta}{\tan \theta} = \frac {\tan \theta}{\tan \theta} + \frac {\cot \theta}{\tan \theta}$

now what? - Sep 14th 2008, 10:02 AMrobasc
On the left side I know that I will be left with tan 0.

but the right side confuses me.

wait I think I got it:

tan 0 = sin 0 / cos 0 - Sep 14th 2008, 10:06 AMJhevon
- Sep 14th 2008, 10:16 AMrobasc
so the answer is tan 0?

- Sep 14th 2008, 10:20 AMJhevon
huh??

lets not lose sight of the objective here. we are proving an identity. we are trying to turn one side into the other. your objective is to turn $\displaystyle \frac {\tan \theta}{\tan \theta} + \frac {\cot \theta}{\tan \theta}$ into $\displaystyle \csc^2 \theta$ through simplification. that will show that the left hand side and right hand side are the same - Sep 14th 2008, 11:18 AM11rdc11
I'm going to work this out for you since it seems you are a little lost and hopefully me working it will help you understand, but I what you to come back and explain the steps I did so you will learn.

$\displaystyle \frac {\tan \theta + \cot \theta}{\tan \theta}$

$\displaystyle =~\frac {\tan \theta}{\tan \theta} + \frac {\cot \theta}{\tan \theta}$

$\displaystyle = 1+ \frac{cos^2\theta}{sin^2\theta}$

$\displaystyle = \frac{sin^2\theta}{sin^2\theta} + \frac{cos^2\theta}{sin^2\theta}$

$\displaystyle =~\frac{1}{sin^2\theta}$

$\displaystyle =~csc^2\theta$ - Sep 14th 2008, 11:20 AMrobasc
ok I have come up with this:

sin/cos + cos/sin

------------------

sin/cos + sin/cos

cos/sin

-------

sin/cos

cos/sin * cos/sin =

cos^2 0

--------

sin^2 0

this is how I am thinking about it? - Sep 14th 2008, 11:27 AMJhevon
evidently you need help with adding fractions (see posts #3, 9 and 11 here). you cannot cancel like that! and you cannot have sin and cos by themselves, write sin(x) and cos(x) or something

furthermore, as i said, you are proving an identity. if you don't end up with something that looks like the other side, you are wrong

Quote:

this is how I am thinking about it?

11rdc11 did the problem for you. make sure you follow what he did - Sep 14th 2008, 11:36 AMrobasc
I was telling you how I was interpreting the problem.

- Sep 14th 2008, 11:38 AM11rdc11
Oops I didn't notice that help canceled out $\displaystyle \frac{sin\theta}{cos\theta}$ like that

- Sep 14th 2008, 11:49 AMJhevon
yes, i know. i was messing with you because you put a question mark at the end.

study what 11rdc11 did and make sure you have no questions. then, make sure you review fractions and trig identities. things like $\displaystyle \sin^2 \theta + \cos^2 \theta = 1$, stuff like that you should know by heart. you are in for a LOT of pain if you do not get these basic things. you don't want to be one of those students that end up in calculus and can't do basic algebra, trust me