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anonymous
 one year ago
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
 one year ago
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))

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sec (0) sec(0)    = 2csc * (0) csc(0)cot(0) csc(0)+cot(0)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.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))

anonymous
 one year ago
Best ResponseYou've already chosen the best response.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)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(2csc*(2theta)+csc(0))(2csc*(2theta)csc(0))  = 2csc*(0) (csc^2(0)cot^2(0))

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0csc^2(0)cot^2(0)=1 ((2csc*(2theta)+csc(0))2csc*(2theta)+csc(0)) 2csc(0)=2csc(0)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0confusion and confusion o,o how did you get 2 there http://prntscr.com/8qb8cf ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im 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
 one year ago
Best ResponseYou've already chosen the best response.0hmm sure i'll just start from http://prntscr.com/8qb8um not gonna start over by different way

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0omg thank you! you are a lifesaver. my teacher already thinks im crazy with some of the answers i give her

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0distribute \[\large\rm \frac{ \sec* \csc + \sec*\cot \sec*\csc + \sec *\cot }{ \csc^2 \theta  \cot^2 \theta }\] ignore theta for a sec

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0what's the definition of sec? like csc = 1/sin so sec = ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0oh 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
 one year ago
Best ResponseYou've already chosen the best response.0\[\large\rm \frac{ \sec* \csc + \sec*\cot \sec*\csc + \sec *\cot }{ \csc^2 \theta  \cot^2 \theta }\] first look at the numerator combine like terms !let me know what u get :=)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0like sec*csc and sec*csc are like terms so sec*csc  sec*csc = ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0that is a different question right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i redid the equation

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0i'm confused don't know how u got tan and cos at the beginning

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0the work u posted here is almost correct but there was a little mistake so i'll say keep the same method

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\[\large\rm \frac{ \sec* \csc + \sec*\cot \sec*\csc + \sec *\cot }{ \csc^2 \theta  \cot^2 \theta }\] just two steps then we are done !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh 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
 one year ago
Best ResponseYou've already chosen the best response.0its 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
 one year ago
Best ResponseYou've already chosen the best response.0wouldnt it cancel eachother out because sec*csc+sec*cot is repeated twice so it would be sec*csc+sec*cot as the numerator

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0yes 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
 one year ago
Best ResponseYou've already chosen the best response.0cot = cos /sin and tan = sin/cos bot hare reciprocal of each other

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0oh well latex doesn't work dw:1444619074859:dw

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0use the identity cot^2 (x) +1 = csc^2 move the cot^2 to the right side what would you get ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\[\huge\rm \frac{ 2(\frac{1}{\cos}*\frac{\cos}{\sin}) }{ \sec^2 \theta \cot^2 }\] 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

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0http://math2.org/math/trig/identities.htm copy these down on ur notes identities gt g

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would i distribute. like 2 x the things in the parentheses

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 could you help with the last bit?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what are we trying to solve. I'm confused.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it the picture you posed?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes the very first picture posted. I have to verify the identity and justify each step

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.0let me just do it by hand from the beginning. one sec :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay 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
 one year ago
Best ResponseYou've already chosen the best response.0hahaa! i'm sure we will get it done!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you lol im going on about 18 hours doing school work today!! _

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok sweet, i got it! lets do this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when 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
 one year ago
Best ResponseYou've already chosen the best response.0good! so lefts ONLY work with the LHS (left hand side)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ \sec(\theta }{ \csc(\theta)\cot(\theta) }\frac{ \sec(\theta) }{ \csc(\theta)+\cot(\theta) }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which would cancel each other out right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now, my approach would be to get the denominators the same.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0see how our denominators are a bit different

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh yea ones () and the others (+)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we can only subtract fractions if there denominators are the same

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0sorry i took too long with syntax

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thants fine im willing to wait to learn

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0pretty much what this is: \[\frac{ A }{ B }+\frac{ C }{ D }=\frac{ AD }{ BD }+\frac{ CB }{ DB }=\frac{ AD+CB }{ BD }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0right, i just want to make sure you are on the right track

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes im doing it right now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now remember the technique of difference of two squares rule: \[(X+Y)(XY)=X^2Y^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so this means, \[\left[ \csc(\theta)\cot(\theta) \right]\left[ \csc(\theta)+\cot(\theta) \right]=\csc ^{2}(\theta)\cot ^{2}(\theta)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0right I was just typing it in now lol it takes me a little longer to write it out on paper

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and thats our denominator right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep! thats our denominator, good!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so lets simplify the big equation that i had posted

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{(\csc(\theta) \cot(\theta))\left[ \csc(\theta)+\cot(\theta) \right] } \frac{ \sec(\theta)\left[ \csc(\theta)\cot(\theta) \right] }{ \left[ \csc(\theta)+\cot(\theta) \right] \left[ \csc(\theta)\cot(\theta) \right]}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay that looks better lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0omg my syntax keeps screwing up

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just had to refresh my page haha a

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{ \csc ^{2}(\theta)\cot ^{2}(\theta) }\frac{ \sec(\theta) \left[ \csc(\theta)\cot(\theta) \right]}{ \csc ^{2}(\theta)\cot ^{2}(\theta) }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so lets group everything now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] \sec(\theta)\left[ \csc(\theta)\cot(\theta) \right]}{ \csc ^{2}(\theta)\cot ^{2}(\theta)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0expand the numerator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ \sec(\theta)\csc(\theta)+\sec(\theta)\cot(\theta)\sec(\theta)\csc(\theta)+\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta)\cot ^{2}(\theta) }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now simplify! the numerator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 2\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta) \cot ^{2}(\theta)}\] this is because sec(theta)csc(theta)sec(theta)csc(theta)=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we are left with this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so t hey each cancel out leaving us with your fraction above?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and thats all that I need to do for the question? It doesnt need to be broken down anymore?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0did nnesha explain to you why \[\csc ^{2}(\theta) \cot ^{2}(\theta)=1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if so i dont quite understand still

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you have a list of trig identities?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://www.purplemath.com/modules/idents.htm where it says \[1 + \cot ^{2}(t) = \csc ^{2}(t)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we just rearrange that equation \[\csc ^{2}(t)  \cot ^{2}(t)=1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so cot^2(t) is always going to = 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no, \[\csc ^{2}(t)  \cot ^{2}(t)=1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh i see what you mean now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0yes because csc^2 (0)cot^2(0)=1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now 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
 one year ago
Best ResponseYou've already chosen the best response.0so \[2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \cos(\theta) } \right)\left( \frac{ \cos(\theta) }{ \sin(\theta) } \right)\] 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
 one year ago
Best ResponseYou've already chosen the best response.0now we know that: \[\csc(\theta)=\frac{ 1 }{ \sin(\theta)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0therefore \[2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \sin(\theta) } \right)=2\csc(\theta)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now we can see that the LHS reduces to \[2\csc(\theta)\] and that means this is also the RHS!!!!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which 2 ces(0) is the final answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0YEP! so that means we are dun!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So for what the teachers asking i simply put everything we have figured out together for her answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0yes that was the first step we did

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Working 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
 one year ago
Best ResponseYou've already chosen the best response.0THANK 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
 one year ago
Best ResponseYou've already chosen the best response.0I gave you best response and a raving testimonial on your page!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hahahaha 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
 one year ago
Best ResponseYou've already chosen the best response.0Im 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
 one year ago
Best ResponseYou've already chosen the best response.0jeez. i live in australia so its 3.30pm here haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im in michigan,usa lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i should be doing lecture writing, but i procrastinate on open study

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ahah jeez. thats hectic

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0see 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

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{blue}{\text{Originally Posted by}}\) @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
 one year ago
Best ResponseYou've already chosen the best response.0and yea sorry *net issues* :/
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