anonymous
  • anonymous
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))
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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anonymous
  • anonymous
sec (0) sec(0) ------------- - ------------- = 2csc * (0) csc(0)-cot(0) csc(0)+cot(0)
anonymous
  • 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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anonymous
  • 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
  • anonymous
(2csc*(2theta)+csc(0))-(2csc*(2theta)-csc(0)) --------------------------------------------- = 2csc*(0) (csc^2(0)-cot^2(0))
anonymous
  • anonymous
csc^2(0)-cot^2(0)=1 ((2csc*(2theta)+csc(0))-2csc*(2theta)+csc(0)) 2csc(0)=2csc(0)
Nnesha
  • Nnesha
confusion and confusion o,o how did you get 2 there http://prntscr.com/8qb8cf ?
anonymous
  • 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
  • Nnesha
hmm sure i'll just start from http://prntscr.com/8qb8um not gonna start over by different way
anonymous
  • anonymous
omg thank you! you are a lifesaver. my teacher already thinks im crazy with some of the answers i give her
Nnesha
  • 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
  • Nnesha
what's the definition of sec? like csc = 1/sin so sec = ?
anonymous
  • anonymous
im not sure:/
Nnesha
  • 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
  • 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
  • Nnesha
like sec*csc and -sec*csc are like terms so sec*csc - sec*csc = ?
anonymous
  • anonymous
Nnesha
  • Nnesha
that is a different question right ?
anonymous
  • anonymous
i redid the equation
Nnesha
  • Nnesha
i'm confused don't know how u got tan and cos at the beginning
Nnesha
  • Nnesha
the work u posted here is almost correct but there was a little mistake so i'll say keep the same method
Nnesha
  • 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
  • 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
  • Nnesha
ohh i was so confused.
anonymous
  • anonymous
my baddd :)
Nnesha
  • 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
  • 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
  • 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
  • Nnesha
cot = cos /sin and tan = sin/cos bot hare reciprocal of each other
Nnesha
  • Nnesha
oh well latex doesn't work |dw:1444619074859:dw|
Nnesha
  • Nnesha
use the identity cot^2 (x) +1 = csc^2 move the cot^2 to the right side what would you get ?
Nnesha
  • 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
  • anonymous
okay
Nnesha
  • Nnesha
http://math2.org/math/trig/identities.htm copy these down on ur notes identities gt g
anonymous
  • anonymous
would i distribute. like 2 x the things in the parentheses
anonymous
  • anonymous
@jim_thompson5910 could you help with the last bit?
anonymous
  • anonymous
@shaik0124
anonymous
  • anonymous
what are we trying to solve. I'm confused.
anonymous
  • anonymous
is it the picture you posed?
anonymous
  • anonymous
posted*
anonymous
  • anonymous
yes the very first picture posted. I have to verify the identity and justify each step
anonymous
  • anonymous
oh right
anonymous
  • 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
  • anonymous
let me just do it by hand from the beginning. one sec :)
anonymous
  • 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
  • anonymous
hahaa! i'm sure we will get it done!
anonymous
  • anonymous
thank you lol im going on about 18 hours doing school work today!! -_-
anonymous
  • anonymous
ok sweet, i got it! lets do this
anonymous
  • anonymous
alright:)
anonymous
  • 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
  • anonymous
yes makes sense
anonymous
  • anonymous
good! so lefts ONLY work with the LHS (left hand side)
anonymous
  • anonymous
\[\frac{ \sec(\theta }{ \csc(\theta)-\cot(\theta) }-\frac{ \sec(\theta) }{ \csc(\theta)+\cot(\theta) }\]
anonymous
  • anonymous
which would cancel each other out right
anonymous
  • anonymous
now, my approach would be to get the denominators the same.
anonymous
  • anonymous
see how our denominators are a bit different
anonymous
  • anonymous
oh yea ones (-) and the others (+)
anonymous
  • anonymous
we can only subtract fractions if there denominators are the same
anonymous
  • 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
  • anonymous
sorry i took too long with syntax
anonymous
  • anonymous
thants fine im willing to wait to learn
anonymous
  • anonymous
pretty much what this is: \[\frac{ A }{ B }+\frac{ C }{ D }=\frac{ AD }{ BD }+\frac{ CB }{ DB }=\frac{ AD+CB }{ BD }\]
anonymous
  • anonymous
right, i just want to make sure you are on the right track
anonymous
  • anonymous
got it?
anonymous
  • anonymous
yes im doing it right now
anonymous
  • anonymous
now remember the technique of difference of two squares rule: \[(X+Y)(X-Y)=X^2-Y^2\]
anonymous
  • anonymous
so this means, \[\left[ \csc(\theta)-\cot(\theta) \right]\left[ \csc(\theta)+\cot(\theta) \right]=\csc ^{2}(\theta)-\cot ^{2}(\theta)\]
anonymous
  • anonymous
right?
anonymous
  • anonymous
right I was just typing it in now lol it takes me a little longer to write it out on paper
anonymous
  • anonymous
and thats our denominator right
anonymous
  • anonymous
yep! thats our denominator, good!
anonymous
  • anonymous
so lets simplify the big equation that i had posted
anonymous
  • anonymous
which one?
anonymous
  • 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
  • anonymous
that one
anonymous
  • anonymous
okay that looks better lol
anonymous
  • anonymous
omg my syntax keeps screwing up
anonymous
  • anonymous
just had to refresh my page haha a
anonymous
  • 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
  • anonymous
good?
anonymous
  • anonymous
so lets group everything now
anonymous
  • 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
  • anonymous
expand the numerator
anonymous
  • 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
  • anonymous
now simplify! the numerator
anonymous
  • anonymous
okay
anonymous
  • 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
  • anonymous
so we are left with this
anonymous
  • anonymous
so t hey each cancel out leaving us with your fraction above?
anonymous
  • anonymous
they*
anonymous
  • anonymous
yep!
anonymous
  • anonymous
and thats all that I need to do for the question? It doesnt need to be broken down anymore?
anonymous
  • anonymous
nope we keep going
anonymous
  • anonymous
okay i thought so
anonymous
  • anonymous
did nnesha explain to you why \[\csc ^{2}(\theta) -\cot ^{2}(\theta)=1\]
anonymous
  • anonymous
if so i dont quite understand still
anonymous
  • anonymous
do you have a list of trig identities?
anonymous
  • anonymous
yes
anonymous
  • anonymous
http://www.purplemath.com/modules/idents.htm where it says \[1 + \cot ^{2}(t) = \csc ^{2}(t)\]
anonymous
  • anonymous
so we just re-arrange that equation \[\csc ^{2}(t) - \cot ^{2}(t)=1\]
anonymous
  • anonymous
so cot^2(t) is always going to = 1
anonymous
  • anonymous
no, \[\csc ^{2}(t) - \cot ^{2}(t)=1\]
anonymous
  • anonymous
oh i see what you mean now
anonymous
  • 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
  • anonymous
right?
anonymous
  • anonymous
correct
anonymous
  • anonymous
awesome :)
anonymous
  • 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
  • anonymous
understand that?
anonymous
  • anonymous
yes because csc^2 (0)-cot^2(0)=1
anonymous
  • 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
  • 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
  • anonymous
now we know that: \[\csc(\theta)=\frac{ 1 }{ \sin(\theta)}\]
anonymous
  • anonymous
therefore \[2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \sin(\theta) } \right)=2\csc(\theta)\]
anonymous
  • anonymous
now we can see that the LHS reduces to \[2\csc(\theta)\] and that means this is also the RHS!!!!!!
anonymous
  • anonymous
which 2 ces(0) is the final answer
anonymous
  • anonymous
YEP! so that means we are dun!
anonymous
  • anonymous
So for what the teachers asking i simply put everything we have figured out together for her answer
anonymous
  • 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
  • anonymous
yes that was the first step we did
anonymous
  • 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
:)
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
  • anonymous
: ) you're too kind
anonymous
  • anonymous
I gave you best response and a raving testimonial on your page!
anonymous
  • 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
  • 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
  • anonymous
jeez. i live in australia so its 3.30pm here haha
anonymous
  • anonymous
im in michigan,usa lol
anonymous
  • anonymous
i should be doing lecture writing, but i procrastinate on open study
anonymous
  • anonymous
ahah jeez. thats hectic
anonymous
  • 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
  • anonymous
haha good night
anonymous
  • anonymous
goodnight(;
Nnesha
  • 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
  • Nnesha
and yea sorry *net issues* :/

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