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sec (0) sec(0)
------------- - ------------- = 2csc * (0)
csc(0)-cot(0) csc(0)+cot(0)

csc^2(0)-cot^2(0)=1
((2csc*(2theta)+csc(0))-2csc*(2theta)+csc(0))
2csc(0)=2csc(0)

confusion and confusion o,o
how did you get 2 there http://prntscr.com/8qb8cf ?

hmm sure
i'll just start from http://prntscr.com/8qb8um
not gonna start over by different way

what's the definition of sec?
like csc = 1/sin
so sec = ?

im not sure:/

like sec*csc and -sec*csc are like terms
so sec*csc - sec*csc = ?

that is a different question right ?

i redid the equation

i'm confused don't know how u got tan and cos at the beginning

ohh i was so confused.

my baddd :)

cot = cos /sin
and tan = sin/cos
bot hare reciprocal of each other

oh well latex doesn't work |dw:1444619074859:dw|

use the identity cot^2 (x) +1 = csc^2
move the cot^2 to the right side
what would you get ?

okay

http://math2.org/math/trig/identities.htm copy these down on ur notes
identities gt g

would i distribute. like 2 x the things in the parentheses

@jim_thompson5910 could you help with the last bit?

what are we trying to solve. I'm confused.

is it the picture you posed?

posted*

yes the very first picture posted. I have to verify the identity and justify each step

oh right

let me just do it by hand from the beginning. one sec :)

hahaa! i'm sure we will get it done!

thank you lol im going on about 18 hours doing school work today!! -_-

ok sweet, i got it! lets do this

alright:)

yes makes sense

good! so lefts ONLY work with the LHS (left hand side)

which would cancel each other out right

now, my approach would be to get the denominators the same.

see how our denominators are a bit different

oh yea ones (-) and the others (+)

we can only subtract fractions if there denominators are the same

sorry i took too long with syntax

thants fine im willing to wait to learn

right, i just want to make sure you are on the right track

got it?

yes im doing it right now

now remember the technique of difference of two squares rule:
\[(X+Y)(X-Y)=X^2-Y^2\]

right?

right I was just typing it in now lol it takes me a little longer to write it out on paper

and thats our denominator right

yep! thats our denominator, good!

so lets simplify the big equation that i had posted

which one?

that one

okay that looks better lol

omg my syntax keeps screwing up

just had to refresh my page haha a

good?

so lets group everything now

expand the numerator

now simplify! the numerator

okay

so we are left with this

so t hey each cancel out leaving us with your fraction above?

they*

yep!

and thats all that I need to do for the question? It doesnt need to be broken down anymore?

nope we keep going

okay i thought so

did nnesha explain to you why \[\csc ^{2}(\theta) -\cot ^{2}(\theta)=1\]

if so i dont quite understand still

do you have a list of trig identities?

yes

http://www.purplemath.com/modules/idents.htm
where it says \[1 + \cot ^{2}(t) = \csc ^{2}(t)\]

so we just re-arrange that equation
\[\csc ^{2}(t) - \cot ^{2}(t)=1\]

so cot^2(t) is always going to = 1

no,
\[\csc ^{2}(t) - \cot ^{2}(t)=1\]

oh i see what you mean now

right?

correct

awesome :)

understand that?

yes because csc^2 (0)-cot^2(0)=1

now we know that:
\[\csc(\theta)=\frac{ 1 }{ \sin(\theta)}\]

therefore
\[2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \sin(\theta) } \right)=2\csc(\theta)\]

now we can see that the LHS reduces to
\[2\csc(\theta)\]
and that means this is also the RHS!!!!!!

which 2 ces(0) is the final answer

YEP! so that means we are dun!

So for what the teachers asking i simply put everything we have figured out together for her answer

yes that was the first step we did

:)

: ) you're too kind

I gave you best response and a raving testimonial on your page!

jeez. i live in australia so its 3.30pm here haha

im in michigan,usa lol

i should be doing lecture writing, but i procrastinate on open study

ahah jeez. thats hectic

haha good night

goodnight(;

and yea sorry *net issues* :/