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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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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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))

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(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)) --------------------------------------------- = 2csc*(0) (csc^2(0)-cot^2(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 ?
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
hmm sure i'll just start from http://prntscr.com/8qb8um not gonna start over by different way
omg thank you! you are a lifesaver. my teacher already thinks im crazy with some of the answers i give her
distribute \[\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }\] ignore theta for a sec
what's the definition of sec? like csc = 1/sin so sec = ?
im not sure:/
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
\[\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }\] first look at the numerator combine like terms !let me know what u get :=)
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
the work u posted here is almost correct but there was a little mistake so i'll say keep the same method
\[\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }\] just two steps then we are done !
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
ohh i was so confused.
my baddd :)
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 :=)
wouldnt it cancel eachother out because sec*csc+sec*cot is repeated twice so it would be sec*csc+sec*cot as the numerator
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
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 ?
\[\huge\rm \frac{ 2(\frac{1}{\cos}*\frac{\cos}{\sin}) }{ \sec^2 \theta -\cot^2 }\] 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
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
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
let me just do it by hand from the beginning. one sec :)
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
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:)
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?
yes makes sense
good! so lefts ONLY work with the LHS (left hand side)
\[\frac{ \sec(\theta }{ \csc(\theta)-\cot(\theta) }-\frac{ \sec(\theta) }{ \csc(\theta)+\cot(\theta) }\]
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
so we are going to multiply the first fraction by the second denominator and vide versa...watch \[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{(\csc(\theta) -\cot(\theta))\left[ \csc(\theta)+\cot(\theta) \right] }- \frac{ \sec(\theta)\left[ \csc(\theta)-\cot(\theta) \right] }{ \left[ \csc(\theta)+\cot(\theta) \right] \left[ \csc(\theta)-\cot(\theta) \right]}\]
sorry i took too long with syntax
thants fine im willing to wait to learn
pretty much what this is: \[\frac{ A }{ B }+\frac{ C }{ D }=\frac{ AD }{ BD }+\frac{ CB }{ DB }=\frac{ AD+CB }{ BD }\]
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: \[(X+Y)(X-Y)=X^2-Y^2\]
so this means, \[\left[ \csc(\theta)-\cot(\theta) \right]\left[ \csc(\theta)+\cot(\theta) \right]=\csc ^{2}(\theta)-\cot ^{2}(\theta)\]
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?
\[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{(\csc(\theta) -\cot(\theta))\left[ \csc(\theta)+\cot(\theta) \right] }- \frac{ \sec(\theta)\left[ \csc(\theta)-\cot(\theta) \right] }{ \left[ \csc(\theta)+\cot(\theta) \right] \left[ \csc(\theta)-\cot(\theta) \right]}\]
that one
okay that looks better lol
omg my syntax keeps screwing up
just had to refresh my page haha a
\[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }-\frac{ \sec(\theta) \left[ \csc(\theta)-\cot(\theta) \right]}{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }\]
good?
so lets group everything now
\[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] -\sec(\theta)\left[ \csc(\theta)-\cot(\theta) \right]}{ \csc ^{2}(\theta)-\cot ^{2}(\theta)}\]
expand the numerator
\[\frac{ \sec(\theta)\csc(\theta)+\sec(\theta)\cot(\theta)-\sec(\theta)\csc(\theta)+\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }\]
now simplify! the numerator
okay
\[\frac{ 2\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta) -\cot ^{2}(\theta)}\] this is because sec(theta)csc(theta)-sec(theta)csc(theta)=0
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
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 \[\csc ^{2}(t) - \cot ^{2}(t)=1\]
right?
correct
awesome :)
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 \[2\sec(\theta)\cot(\theta) \]
understand that?
yes because csc^2 (0)-cot^2(0)=1
now all we need to know is some basic trig formulas: \[\csc(\theta)=\frac{ 1 }{ \sin (\theta) }\] \[\sec(\theta)=\frac{ 1 }{ \cos(\theta) }\] and \[\cot(\theta)=\frac{ 1 }{ \tan(\theta) }=\frac{ \cos(\theta) }{ \sin(\theta) }\]
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: \[2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \sin(\theta) } \right)\]
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
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
yes that was the first step we did
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)
:)
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!!!
: ) you're too kind
I gave you best response and a raving testimonial on your page!
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!
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)
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
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
haha good night
goodnight(;
\(\color{blue}{\text{Originally Posted by}}\) @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
and yea sorry *net issues* :/

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