SOLUTION: Please help me verify this equation: cscx - cscx ------ ------ =2sec^2x 1+cscx 1-cscx yes, th

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Question 173390: Please help me verify this equation:
cscx - cscx
------ ------ =2sec^2x
1+cscx 1-cscx

yes, the cscx's are over 1+cscx and 1-cscx. Those are fractions.
So far i have gotten stuck on this porblem. i have expanded csc x into 1/sinx in both parts of the left side of the equation which gets me:
1/sinx - 1/sinx
-------- --------- = 2sec^2x
1+1/sinx 1-1/sinx

After that i tried to achieve a common denominator on the left side which is where i get lost. I'm not sure what step to do after that or how to solve it.


Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
good one.
maybe i got the answer.
you want to prove this equation is true.
here goes:
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let c = csc(x)
let s = sec(x)
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your equation becomes:
c%2F%281%2Bc%29+-+c%2F%281-c%29+=+2s%5E2
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if you multiply both sides of the equation by (1-c)*(1+c), you get:
c%2A%281-c%29+-+c%2A%281%2Bc%29+=+2s%5E2%2A%281-c%29%2A%281%2Bc%29
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simplifying, this becomes:
c+-+c%5E2+-+c+%2B+c%5E2+=+2s%5E2+%2A+%281-c%5E2%29
which becomes:
-2c%5E2+=+2s%5E2+%2A+%281-c%5E2%29
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this is where substitutions come in.
you know that:
sin%5E2%28x%29+%2B+cos%5E2%28x%29+=+1
you also know that:
sin(x) = 1/csc(x)
and that:
cos(x) = 1/sec(x)
substituting for sin(x) and cos(x), the equation becomes:
1%2Fcsc%5E2%28x%29+%2B+1%2Fsec%5E2%28x%29+=+1
if you multiply both sides of this equation by csc%5E2%28x%29%2Asec%5E2%28x%29, the equation becomes:
sec%5E2%28x%29+%2B+csc%5E2%28x%29+=+sec%5E2%28x%29%2Acsc%5E2%28x%29
subtract csc%5E2%28x%29 from both sides of this equation to get:
sec%5E2%28x%29+=+sec%5E2%28x%29%2Acsc%5E2%28x%29+-+csc%5E%28x%29
simplify this on the right hand side to become:
sec%5E2%28x%29+=+csc%5E2%28x%29+%2A+%28sec%5E2%28x%29+-+1%29
divide both sides of this equation by %28sec%5E2%28x%29+-+1%29 to get:
sec%5E2%28x%29%2F%28sec%5E2%28x%29+-+1%29+=+csc%5E2%28x%29
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this is the same as:
csc%5E2%28x%29+=+sec%5E2%28x%29%2F%28sec%5E2%28x%29+-+1%29
and, since we originally let c = csc(x) and we let s = sec(x), this equation now becomes:
c%5E2+=+s%5E2%2F%28s%5E2-1%29
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we can now substitute for c^2 in the original equation we derived above, which was:
-2c%5E2+=+2s%5E2+%2A+%281-c%5E2%29
simplifying this, it becomes:
-2c%5E2+=+2s%5E2+-+2s%5E2%2Ac%5E2
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substituting for c%5E2, that equation becomes:
-2%2A%28s%5E2%2F%28s%5E2-1%29%29+=+2s%5E2+-+2s%5E2%2As%5E2%2F%28s%5E2-1%29
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if we multiply both sides of this equation by s%5E2-1, we get:
-2s%5E2+=+2s%5E2+%2A+%28s%5E2-1%29+-+2s%5E2%2As%5E2
this becomes:
-2s%5E2+=+2s%5E4+-+2s%5E2+-+2s%5E4
which becomes:
-2s%5E2+=+-2s%5E2
if we multiply both sides of this equation by (-1), it becomes:
2s%5E2+=+2s%5E2
which proves that the original equation of:
c%2F%281%2Bc%29+-+c%2F%281-c%29+=+2s%5E2
is true.
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since we originally substituted c for csc(x), and s for sec(x), the original equation becomes:
csc%28x%29%2F%281%2Bcsc%28x%29%29+-+csc%28x%29%2F%281-csc%28x%29%29+=+2sec%5E2%28x%29
and the proven identify becomes:
2sec%5E2%28x%29+=+2sec%5E2%28x%29
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