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(csc x + 1)(csc x - 1)

Mathematics
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FOIL it and apply trig identity
so its \[(\csc x)^2-1=cot^22x\]
yep

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Other answers:

@itsmylife how did you get that
trig identity
how
can you show how you got that plz
pythagorean identity
There are three Pythagorean trig identities that you simply must memorize.
go here @pvs285 http://www.sosmath.com/trig/Trig5/trig5/trig5.html
can you just show the work to answer this question plz?
\[\left( \csc x+1 \right)\left( \csc x-1 \right)=\csc ^{2}x-1=\cot ^{2}x\]
\[\sin^2x+\cos^2x=1,\frac{ \sin^2 x}{ \sin^2x }+\frac{ \cos^2x }{ \sin^2x }=\frac{ 1 }{ \sin^2x }, 1+\cot^2x=cosec^2x\]
It is against the policies of this site to simply provide answers.
huh u said it was cot^2
ik i whant you to explain it to me so i uderstand
it might be a mistake ;)
huh
pls help
plz
\[\csc ^{2}x-\cot ^{2}=1,or \csc ^{2}x-1=\cot ^{2}x\] It is an identity.
so there is no work involved?
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.
$$ \tt \bf \Large(\csc x + 1)(\csc x -1)\\ \Large=\csc^2x-\csc x +\csc x -1\\ \Large=\csc^2x-1\\\\ \Large=\left(\dfrac{1}{\sin x}\right)^2-1\\ \Large=\dfrac{1}{\sin^2x}-1\\ \Large=\dfrac{1}{\sin^2x}-\dfrac{\sin^2x}{\sin^2x}\\\\ \Large=\dfrac{1-\sin^2x}{\cos^2x}\\\\ \Large=\dfrac{\cos^2x}{\sin^2x}\\\\ \Large=\cot^2x $$ Please let me kow if you have any questions. Good luck.

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