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prove using identities that ((sec^2)x)/(((sec^2)x)-1)=(csc^2)x

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Okay, so let me try this: did you mean: \[Sec^2(x)/(Sec^2x - 1) = Csc^2(x)\]
yea. sorry i don't know how get it like that on my computer
okay so now workin on it

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Other answers:

thank you so much
There are a few Sec and Csc identities, but I never remember them so im just gonna convert em to Sin and Cos: Sec(x) = 1/Cos(x) Csc(x) = 1/Sin(x)
\[[1/Cos^2x] / [ (1/Cos^2x)-1 ] = 1/Sin^2x\]
Now let's simplify this part: \[(1/Cos^2x) - 1\] = \[(1-Cos^2x)/Cos^2x\]
wait. how did you do that?
find the common denominator
1 = Cos^2 / Cos^2... get it?
oh . yea, sorry. i got it
So now you have to divide 1/Cos^2x by (1-Cos^2x)/Cos^2x which is the same as multiplying by its inverse
\[[1/Cos^2x] * [Cos^2x/(1-Cos^2x)] = 1/Sin^2x\]
Cancel out cos^2s
\[1/(1-Cos^2x) = 1/Sin^2x\]
Cross multiply:
wait. can't you just say 1 / (sin^2x) =csc^2x
and then be done? or is that skipping a step?
yeah, but i prefer keeping everything in sin and cosine
bcuz its easy to simplify
\[Sin^2x = 1 - Cos^2x\]
See that's proved. Sin^2x + Cos^2x = 1!!!!
get it?
yes. thank you so much!
k off to helpin someone else.. any questions: email me and if u liked my explanation, bcum a fan YAY!
deff will. that was so helpful. thanks again!

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