Can someone check my answer, I feel like I did it completely wrong!!
13. Verify the identity. Justify each step.
sec (0) sec(0)
------------- - ------------- = 2csc * (0)
csc(0)-cot(0) csc(0)+cot(0)
(sec(0)*(csc(0)+ cot(0)) - (sec(0)*(csc(0)-cot(0)))
--------------------------------------------------- = 2csc*(0)
(csc(0)-cot(0))*(csc(0)*(csc(0)+cot(0))
(sec(0)*csc(0)+cot(0)*sec(0))-(2sec(0)*csc(0)-cot(0)*sec(0))
------------------------------------------------------------
(csc^2(0)-cot^2(0))
=2csc*(0)
(2csc*(2theta)+csc(0))-(2csc*(2theta)-csc(0))

- anonymous

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- anonymous

##### 1 Attachment

- anonymous

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

- anonymous

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

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## More answers

- anonymous

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

- anonymous

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

- anonymous

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

- Nnesha

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

- anonymous

im not quite sure. I watched a video about how to do it but i feel like i didnt learn a thing lol could you help me

- Nnesha

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

- anonymous

omg thank you! you are a lifesaver. my teacher already thinks im crazy with some of the answers i give her

- Nnesha

distribute \[\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }\]
ignore theta for a sec

- Nnesha

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

- anonymous

im not sure:/

- Nnesha

oh okay thats fine
but these are very important \[\sin \theta= \frac{1}{\csc \theta} ~~~\cos \theta =\frac{1}{\sec \theta }~~~ \tan \theta= \frac{1}{\cot \theta }\]
sin cos tan and csc sec cot are reciprocal

- Nnesha

\[\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }\]
first look at the numerator
combine like terms !let me know what u get :=)

- Nnesha

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

- anonymous

##### 1 Attachment

- Nnesha

that is a different question right ?

- anonymous

i redid the equation

- Nnesha

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

- Nnesha

the work u posted here is almost correct but there was a little mistake so i'll say keep the same method

- Nnesha

\[\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }\]
just two steps
then we are done !

- anonymous

oh shoot it was the wrong one im sorry. i guess ive been doing too much work today apparently lol ill just start from where you lfet off then

- Nnesha

ohh i was so confused.

- anonymous

my baddd :)

- Nnesha

its okay \[\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }\]
can you simplify the numerator just like simple algebra :=) combine like terms :=)

- anonymous

wouldnt it cancel eachother out because sec*csc+sec*cot is repeated twice
so it would be
sec*csc+sec*cot as the numerator

- Nnesha

yes right sec*cot + sec *cot = 2 sec*cot
\[\huge\rm \frac{ 2\sec*\cot }{ \sec^2 - \cot^2}\]
we can convert sec to 1/cos and cot to cos/sin

- Nnesha

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

- Nnesha

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

- Nnesha

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

- Nnesha

\[\huge\rm \frac{ 2(\frac{1}{\cos}*\frac{\cos}{\sin}) }{ \sec^2 \theta -\cot^2 }\]
simplify the numerator
if we solve cot^2 theta +1 =csc^2 for 1
we would get csc^2 -cot^2 =1
so you can replace `csc^2 -cot^2` with 1
simplify the numerator

- anonymous

okay

- Nnesha

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

- anonymous

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

- anonymous

@jim_thompson5910 could you help with the last bit?

- anonymous

@shaik0124

- anonymous

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

- anonymous

is it the picture you posed?

- anonymous

posted*

- anonymous

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

- anonymous

oh right

- anonymous

i was told that i had it mostly right. we were working from the second line or the third post but i think she started having connection issues

- anonymous

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

- anonymous

okay thank you so much! this is the last problem that I have to fix and weve been working on it for probably an hour

- anonymous

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

- anonymous

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

- anonymous

ok sweet, i got it! lets do this

- anonymous

alright:)

- anonymous

when i approach these question, what i tend to do is solve the massive weird stuff and compare it to the side which is simplified.
so in other words, what i would do with this question is to simplify the left hand side and compare it to the right hand side, right?

- anonymous

yes makes sense

- anonymous

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

- anonymous

\[\frac{ \sec(\theta }{ \csc(\theta)-\cot(\theta) }-\frac{ \sec(\theta) }{ \csc(\theta)+\cot(\theta) }\]

- anonymous

which would cancel each other out right

- anonymous

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

- anonymous

see how our denominators are a bit different

- anonymous

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

- anonymous

we can only subtract fractions if there denominators are the same

- anonymous

so we are going to multiply the first fraction by the second denominator and vide versa...watch
\[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{(\csc(\theta) -\cot(\theta))\left[ \csc(\theta)+\cot(\theta) \right] }- \frac{ \sec(\theta)\left[ \csc(\theta)-\cot(\theta) \right] }{ \left[ \csc(\theta)+\cot(\theta) \right] \left[ \csc(\theta)-\cot(\theta) \right]}\]

- anonymous

sorry i took too long with syntax

- anonymous

thants fine im willing to wait to learn

- anonymous

pretty much what this is:
\[\frac{ A }{ B }+\frac{ C }{ D }=\frac{ AD }{ BD }+\frac{ CB }{ DB }=\frac{ AD+CB }{ BD }\]

- anonymous

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

- anonymous

got it?

- anonymous

yes im doing it right now

- anonymous

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

- anonymous

so this means,
\[\left[ \csc(\theta)-\cot(\theta) \right]\left[ \csc(\theta)+\cot(\theta) \right]=\csc ^{2}(\theta)-\cot ^{2}(\theta)\]

- anonymous

right?

- anonymous

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

- anonymous

and thats our denominator right

- anonymous

yep! thats our denominator, good!

- anonymous

so lets simplify the big equation that i had posted

- anonymous

which one?

- anonymous

\[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{(\csc(\theta) -\cot(\theta))\left[ \csc(\theta)+\cot(\theta) \right] }- \frac{ \sec(\theta)\left[ \csc(\theta)-\cot(\theta) \right] }{ \left[ \csc(\theta)+\cot(\theta) \right] \left[ \csc(\theta)-\cot(\theta) \right]}\]

- anonymous

that one

- anonymous

okay that looks better lol

- anonymous

omg my syntax keeps screwing up

- anonymous

just had to refresh my page haha a

- anonymous

\[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }-\frac{ \sec(\theta) \left[ \csc(\theta)-\cot(\theta) \right]}{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }\]

- anonymous

good?

- anonymous

so lets group everything now

- anonymous

\[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] -\sec(\theta)\left[ \csc(\theta)-\cot(\theta) \right]}{ \csc ^{2}(\theta)-\cot ^{2}(\theta)}\]

- anonymous

expand the numerator

- anonymous

\[\frac{ \sec(\theta)\csc(\theta)+\sec(\theta)\cot(\theta)-\sec(\theta)\csc(\theta)+\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }\]

- anonymous

now simplify! the numerator

- anonymous

okay

- anonymous

\[\frac{ 2\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta) -\cot ^{2}(\theta)}\]
this is because sec(theta)csc(theta)-sec(theta)csc(theta)=0

- anonymous

so we are left with this

- anonymous

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

- anonymous

they*

- anonymous

yep!

- anonymous

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

- anonymous

nope we keep going

- anonymous

okay i thought so

- anonymous

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

- anonymous

if so i dont quite understand still

- anonymous

do you have a list of trig identities?

- anonymous

yes

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

oh i see what you mean now

- anonymous

so where we how we have our denominator as \[\csc ^{2}(t) - \cot ^{2}(t)\]
we can just substitute that as 1 since we know the trig identity that
\[\csc ^{2}(t) - \cot ^{2}(t)=1\]

- anonymous

right?

- anonymous

correct

- anonymous

awesome :)

- anonymous

so that means
\[\frac{ 2\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta) -\cot ^{2}(\theta)}\]
REDUCES to:
\[\frac{ 2\sec(\theta)\cot(\theta) }{ 1}\]
which is simply
\[2\sec(\theta)\cot(\theta) \]

- anonymous

understand that?

- anonymous

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

- anonymous

now all we need to know is some basic trig formulas:
\[\csc(\theta)=\frac{ 1 }{ \sin (\theta) }\]
\[\sec(\theta)=\frac{ 1 }{ \cos(\theta) }\]
and
\[\cot(\theta)=\frac{ 1 }{ \tan(\theta) }=\frac{ \cos(\theta) }{ \sin(\theta) }\]

- anonymous

so
\[2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \cos(\theta) } \right)\left( \frac{ \cos(\theta) }{ \sin(\theta) } \right)\]
the cos(X) part cancels since we are multiplying fractions, so we are left with:
\[2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \sin(\theta) } \right)\]

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

which 2 ces(0) is the final answer

- anonymous

YEP! so that means we are dun!

- anonymous

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

- anonymous

so what we did was reduce the left hand side right?
and we found that the left hand side equalled to the right hand side

- anonymous

yes that was the first step we did

- anonymous

Working with the LHS, we want to reduce this to see if it equals the RHS.
LHS IMPLIES:
\[\frac{ \sec(\theta }{ \csc(\theta)-\cot(\theta) }-\frac{ \sec(\theta) }{ \csc(\theta)+\cot(\theta) }\]
Getting denominators the same:
\[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{(\csc(\theta) -\cot(\theta))\left[ \csc(\theta)+\cot(\theta) \right] }- \frac{ \sec(\theta)\left[ \csc(\theta)-\cot(\theta) \right] }{ \left[ \csc(\theta)+\cot(\theta) \right] \left[ \csc(\theta)-\cot(\theta) \right]}\]
Now from denominators, we can use the difference of two squares:
\[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }-\frac{ \sec(\theta) \left[ \csc(\theta)-\cot(\theta) \right]}{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }\]
Now subtract the two fractions since we have the denominators the same: I.e get it into one fraction.
\[\frac{ \sec(\theta)\csc(\theta)+\sec(\theta)\cot(\theta)-\sec(\theta)\csc(\theta)+\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }\]
Simplifying the numerator we get:
\[\frac{ 2\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta) -\cot ^{2}(\theta)}\]
Now, using trig identities for the numerator where \[\csc ^{2}(\theta) -\cot ^{2}(\theta)=1\]
Therefore, it reduces to
\[\frac{ 2\sec(\theta)\cot(\theta) }{ 1}\]
Using basic trig identities, we can simplify in terms of sin(x) and cos(x)
\[2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \cos(\theta) } \right)\left( \frac{ \cos(\theta) }{ \sin(\theta) } \right)\]
Cancelling out the cos(theta) we get,
\[2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \sin(\theta) } \right)\]
finally we know
\[\csc(\theta)=\frac{ 1 }{ \sin(\theta)}\]
therefore
\[2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \sin(\theta) } \right)=2\csc(\theta)\]
=RHS (As required)

- anonymous

:)

- anonymous

THANK YOU SO MUCH!!! You are a huge lifesaver!!! This was my last question I had to correct and understand for me to complete my diploma. You truly deserve best response for your very detailed help! Thank you thank you!!!

- anonymous

: ) you're too kind

- anonymous

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

- anonymous

hahahaha if you want you can fan me so if you stuck with trig equations, you can easily find me if i'm online :) well done though.
they are a bit tricky, but there is always a systematic approach!

- anonymous

Im already a fan(; If i have any other questions I will be sure to ask you! thank you again and have a wonderful night/morning (its 1am here)

- anonymous

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

- anonymous

im in michigan,usa lol

- anonymous

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

- anonymous

ahah jeez. thats hectic

- anonymous

see im good at writting so if you need help with THAT let me know lol I will be on tomorrow but right now..im off to bed! my two year old wakes up too early lol

- anonymous

haha good night

- anonymous

goodnight(;

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @Nnesha
\[\huge\rm \frac{ 2(\frac{1}{\cos}*\frac{\cos}{\sin}) }{ \sec^2 \theta -\cot^2 }\]
simplify the numerator
if we solve cot^2 theta +1 =csc^2 for 1
we would get csc^2 -cot^2 =1
so you can replace `csc^2 -cot^2` with 1
simplify the numerator
\(\color{blue}{\text{End of Quote}}\)
we replace sec with 1/sec
and cot with cos/sin
se cancel out the cso theta
\[\huge\rm \frac{ 2(\frac{1}{\cancel{cos}}*\frac{\cancel{cos}}{\sin}) }{ \sec^2 \theta -\cot^2 }\]
and csc =1/sin
like i said they are reciprocal
as mentioned above sec^2 +1 = cot^2
set it equal to one so we can replace sec^2 -cot^2 with 1

- Nnesha

and yea sorry *net issues* :/

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