Expert: Sherman D. - 12/18/2005

Question
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Followup To
Question -
1-prove the identity Tan square=csc square/csc square-1  than -1

2- find the inverse x+1/2-3x

3prove identity  2tan square +3 sec /sec+2 =        2-cos/cos
Answer -
1.)
What do you mean by
(csc²A) - 1 in the denominator and separtely is negavitive 1

2.)
By this do you mean
(x + 1)/(2 - 3x)  yes

3.)
By this do you mean
(2tan²A + 3secA)/((secA) + 2) = (2 - cosA)/cosA

Whereas the A stands for the degree measure.  yes this is the question

Answer
1.) when you have csc square/(csc square - 1)

you need to have a degrees of some sort, the letter "A" or "x" is used most often to represent a degree.

² means squared

(cscA)² is the same as csc²A

go to www.oakroadsystems.com/twt/sixfunc.htm to see what i mean.

also still what do you mean by "THAN"

Are you saying maybe

Tan²A = ((csc²A)/(csc²A - 1)) - 1

If so, here is what i can tell you

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Tan²A = ((1/sin²A)/(((1/(sin²A)) - 1)) - 1

Tan²A = ((1/sin²A)/((1 - sin²A)/sin²A) - 1

Tan²A = ((1/sin²A)*((sin²A)/(1 - sin²A)) - 1

Tan²A = ((sin²A)/((sin²A)(1 - sin²A))) - 1

Tan²A = (1/(1 - sin²A)) - 1

Tan²A = (1 - (1 - sin²A))/(1 - sin²A)

Tan²A = (1 - 1 + sin²A)/(1 - sin²A)

Tan²A = (sin²A)/(1 - sin²A)

1 - sin²A = cos²A

Tan²A = (sin²A)/(cos²A)

same as

Tan²A = (sinA/cosA)²

so as you can see

Tan²A = Tan²A

on 1 - sin²A = cos²A, go to www.math.com/tables/trig/identities.htm

They have sin²A + cos²A = 1, but if you do the math, thats the same as saying cos²A = 1 - sin²A

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2.)
If you have

y = (x + 1)/(2 - 3x)

y = (x + 1)/(-3x + 2)

y = (x + 1)/(-(3x - 2))

y = -(x + 1)/(3x - 2)

swith the "x" and "y"

x = -(y + 1)/(3y - 2)

divide both sides by -1

-x = (y + 1)/(3y - 2)

multiply both sides by (3y - 2)

-3xy + 2x = y + 1

subtract "y" and "2x" from both sides

-3xy - y = -2x + 1

factor the left side

y(-3x + 1) = -2x + 1

divide both sides by (-3x + 1)

y = (-2x + 1)/(-3x + 1)

y = (-(2x - 1))/(-(3x - 1))

y = (2x - 1)/(3x - 1)

I believe that is the answer you are looking for.

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3.)
(2tan²A + 3secA)/(secA + 2) = (2 - cosA)/(cosA)

(2(sin²A/cos²A) + 3(1/cosA))/((1/cosA) + 2) = (2 - cosA)/(cosA)

((2sin²A + 3cosA)/(cos²A))/((1 + 2cosA)/cosA) = (2 - cosA)/(cosA)

((2sin²A + 3cosA)/(cos²A))*((cosA)/(1 + 2cosA)) = (2 - cosA)/(cosA))

(cosA(2sin²A + 3cosA))/((cos²A)(1 + 2cosA)) = (2 - cosA)/(cosA)

(2sin²A + 3cosA)/(cosA(1 + 2cosA)) = (2 - cosA)/(cosA)

sin²A = 1 - cos²A

(2(1 - cos²A) + 3cosA)/(cosA(1 + 2cosA)) = (2 - cosA)/(cosA)

(2 - 2cos²A + 3cosA)/(cosA(1 + 2cosA)) = (2 - cosA)/(cosA)

(-2cos²A + 3cosA + 2)/(cosA(1 + 2cosA)) = (2 - cosA)/(cosA)

(-(2cos²A - 3cosA - 2))/(cosA(1 + 2cosA))

(-(2cosA + 1)(cosA - 2))/(cosA(1 + 2cosA))

(-(1 + 2cosA)(-2 + cosA))/(cosA(1 + 2cosA))

(-(1 + 2cosA)(-(2 - cosA)))/(cosA(1 + 2cosA))

((1 + 2cosA)(2 - cosA))/(cosA(1 + 2cosA))

As you can see, since there is a (1 + 2cosA) on top and bottom, they cancel out, which leaves you with

(2 - cosA)/(cosA) = (2 - cosA)/(cosA)

So you don't get confused, i separated my work from each other, and i also separated the problems from each other.