anonymous
  • anonymous
can somebody please help me with this question? i am completely lost. Use basic identities to simplify the expression. (picture below)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \frac{ \cos^2\theta}{\sin^2\theta}+\csc\theta\sin\theta }\)
SolomonZelman
  • SolomonZelman
some basic identities: \(\large \color{black}{\sin\theta\csc\theta=1}\) \(\large \color{black}{\cos\theta\sec\theta=1}\) \(\large \color{black}{\tan\theta\cot\theta=1}\)

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SolomonZelman
  • SolomonZelman
i still want to ask, do you have any options?
anonymous
  • anonymous
these are my options
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SolomonZelman
  • SolomonZelman
ok, then you need 2 more rules" \(\color{black}{\large \dfrac{\sin\theta}{\cos\theta}=\tan\theta}\) \(\large \color{black}{\tan^2\theta+1=\sec^2\theta}\)
SolomonZelman
  • SolomonZelman
well, the first rule also applies when sine and cosine are raised to any power of n, such that: \(\color{black}{\large \dfrac{\sin^n\theta}{\cos^n\theta}=\tan^n\theta}\) (for all n, of course)
Nnesha
  • Nnesha
or maybe u need 2 More rules :3 Mo\[\huge\rm \frac{ \cos \theta }{ \sin \theta}= \cot \theta \] \[\huge\rm cot^2 \theta +1=\csc^2\theta \]
SolomonZelman
  • SolomonZelman
oh, Nnesha, you are right, because it is cosine/sine, not the other way around as I thought for some reason:) and thus, the 2nd rule you posted comes about.
anonymous
  • anonymous
so, what exactly do i do with the rules you guys gave me?
SolomonZelman
  • SolomonZelman
\(\LARGE \color{blue}{\unicode{x10009999999}}\)\(\LARGE \color{blue}{\unicode{x120D999}}\)\(\LARGE \color{blue}{\unicode{x3D5}}\)\(\LARGE \color{blue}{\unicode{x8594}}\) \(\LARGE \color{blue}{:)}\)
SolomonZelman
  • SolomonZelman
u apply them to your problem
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \frac{ \cos^2\theta}{\sin^2\theta}+\csc\theta\sin\theta }\) apply to the second term, one of the rules that I posted in the very beginning.
anonymous
  • anonymous
okay what i think i understand is cos^2(theta) / sin^2(theta) = 1
anonymous
  • anonymous
then cos/sin = cot
anonymous
  • anonymous
would it then be cot^2(theta) = 1 ?
Nnesha
  • Nnesha
nope.
Nnesha
  • Nnesha
cos^2/sin^2 is not equal to one if cos /sin =cot then cos^2 / sin^2 = cot^2
anonymous
  • anonymous
cot^2 (theta) + 1 ?
Nnesha
  • Nnesha
yes right cot^2 theta +1 which is e qual to what ?
anonymous
  • anonymous
csc^2 (theta) ?
Nnesha
  • Nnesha
yep right
anonymous
  • anonymous
thank you so much!
Nnesha
  • Nnesha
my pleasure

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