anonymous
  • anonymous
(csc x + 1)(csc x - 1)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
NoelGreco
  • NoelGreco
FOIL it and apply trig identity
anonymous
  • anonymous
so its \[(\csc x)^2-1=cot^22x\]
NoelGreco
  • NoelGreco
yep

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anonymous
  • anonymous
@itsmylife how did you get that
anonymous
  • anonymous
trig identity
anonymous
  • anonymous
how
anonymous
  • anonymous
can you show how you got that plz
anonymous
  • anonymous
pythagorean identity
NoelGreco
  • NoelGreco
There are three Pythagorean trig identities that you simply must memorize.
anonymous
  • anonymous
go here @pvs285 http://www.sosmath.com/trig/Trig5/trig5/trig5.html
anonymous
  • anonymous
can you just show the work to answer this question plz?
anonymous
  • anonymous
\[\left( \csc x+1 \right)\left( \csc x-1 \right)=\csc ^{2}x-1=\cot ^{2}x\]
anonymous
  • anonymous
\[\sin^2x+\cos^2x=1,\frac{ \sin^2 x}{ \sin^2x }+\frac{ \cos^2x }{ \sin^2x }=\frac{ 1 }{ \sin^2x }, 1+\cot^2x=cosec^2x\]
NoelGreco
  • NoelGreco
It is against the policies of this site to simply provide answers.
anonymous
  • anonymous
huh u said it was cot^2
anonymous
  • anonymous
ik i whant you to explain it to me so i uderstand
anonymous
  • anonymous
it might be a mistake ;)
anonymous
  • anonymous
huh
anonymous
  • anonymous
pls help
anonymous
  • anonymous
plz
anonymous
  • anonymous
\[\csc ^{2}x-\cot ^{2}=1,or \csc ^{2}x-1=\cot ^{2}x\] It is an identity.
anonymous
  • anonymous
so there is no work involved?
NoelGreco
  • NoelGreco
pvs: it appears you are in over your head. If you have a question such as this, and can't pick up on the idea of a trig identity you might want to slow down and back up.
ybarrap
  • ybarrap
$$ \tt \bf \Large(\csc x + 1)(\csc x -1)\\ \Large=\csc^2x-\csc x +\csc x -1\\ \Large=\csc^2x-1\\\\ \Large=\left(\dfrac{1}{\sin x}\right)^2-1\\ \Large=\dfrac{1}{\sin^2x}-1\\ \Large=\dfrac{1}{\sin^2x}-\dfrac{\sin^2x}{\sin^2x}\\\\ \Large=\dfrac{1-\sin^2x}{\cos^2x}\\\\ \Large=\dfrac{\cos^2x}{\sin^2x}\\\\ \Large=\cot^2x $$ Please let me kow if you have any questions. Good luck.

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