pvs285
(csc x + 1)(csc x - 1)
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NoelGreco
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FOIL it and apply trig identity
itsmylife
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so its \[(\csc x)^2-1=cot^22x\]
NoelGreco
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yep
pvs285
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@itsmylife how did you get that
itsmylife
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trig identity
pvs285
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how
pvs285
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can you show how you got that plz
itsmylife
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pythagorean identity
NoelGreco
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There are three Pythagorean trig identities that you simply must memorize.
pvs285
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can you just show the work to answer this question plz?
surjithayer
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\[\left( \csc x+1 \right)\left( \csc x-1 \right)=\csc ^{2}x-1=\cot ^{2}x\]
itsmylife
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\[\sin^2x+\cos^2x=1,\frac{ \sin^2 x}{ \sin^2x }+\frac{ \cos^2x }{ \sin^2x }=\frac{ 1 }{ \sin^2x }, 1+\cot^2x=cosec^2x\]
NoelGreco
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It is against the policies of this site to simply provide answers.
pvs285
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huh u said it was cot^2
pvs285
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ik i whant you to explain it to me so i uderstand
itsmylife
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it might be a mistake ;)
pvs285
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huh
pvs285
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pls help
pvs285
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plz
surjithayer
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\[\csc ^{2}x-\cot ^{2}=1,or \csc ^{2}x-1=\cot ^{2}x\]
It is an identity.
pvs285
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so there is no work involved?
NoelGreco
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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
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$$
\tt \bf
\Large(\csc x + 1)(\csc x -1)\\
\Large=\csc^2x-\csc x +\csc x -1\\
\Large=\csc^2x-1\\\\
\Large=\left(\dfrac{1}{\sin x}\right)^2-1\\
\Large=\dfrac{1}{\sin^2x}-1\\
\Large=\dfrac{1}{\sin^2x}-\dfrac{\sin^2x}{\sin^2x}\\\\
\Large=\dfrac{1-\sin^2x}{\cos^2x}\\\\
\Large=\dfrac{\cos^2x}{\sin^2x}\\\\
\Large=\cot^2x
$$
Please let me kow if you have any questions.
Good luck.