Proving: (sec^2-6*tan+7)/sec^2-5 = (tan-4)/tan+2

[duffman868]

ok i have been working on this problem for a little bit and am stumped. i need to prove.
(sec^2-6*tan+7)/sec^2-5 = (tan-4)/tan+2

this is what i have done skipping a few steps
((1/u^2)-6*(v/u)+7)/(1-5u^2)/u^2

then (u-6u^3*vu^2*1-5u^3)/u^3 got this by finding a common denominator and flipping the bottom fraction. i missed quite a bit of class just trying to catch up but have no idea what i am doing

 

[d_leet]

ok i have been working on this problem for a little bit and am stumped. i need to prove.
(sec^2-6*tan+7)/sec^2-5 = (tan-4)/tan+2

this is what i have done skipping a few steps
((1/u^2)-6*(v/u)+7)/(1-5u^2)/u^2

then (u-6u^3*vu^2*1-5u^3)/u^3 got this by finding a common denominator and flipping the bottom fraction. i missed quite a bit of class just trying to catch up but have no idea what i am doing

Well firstly why are you making substitutions like that, there is absolutely no reason too and makes it incredibly unobvious when you use a trig identity if you do that. Second your notation is mostly gibberish to me, I have no clue what "tan-4) or "tan+2" are supposed to be.

 

[duffman868]

that was the way i was taught.
sec= secant=1/u
tan= tangent=u/v

so tan-4 would be (u/v)- 2

 

[VietDao29]

ok i have been working on this problem for a little bit and am stumped. i need to prove.
(sec^2-6*tan+7)/sec^2-5 = (tan-4)/tan+2

this is what i have done skipping a few steps
((1/u^2)-6*(v/u)+7)/(1-5u^2)/u^2

then (u-6u^3*vu^2*1-5u^3)/u^3 got this by finding a common denominator and flipping the bottom fraction. i missed quite a bit of class just trying to catch up but have no idea what i am doing

As d_leet already said, tan + 4, tan - 2 really make no sense.
You should write tan(x) + 4, or tan(x) - 2 instead.
back to the problem, I hope you mean:
Prove:
$$\frac{\sec ^ 2 (x) - 6 \tan (x) + 7}{\sec ^ 2 (x) - 5} = \frac{\tan (x) - 4}{\tan (x) + 2}$$
Is that what you mean?
I'll give you a hint, try to change all secant function to tangent function. Do you know:
$$1 + \tan ^ 2 (x) = \sec ^ 2 (x)$$?
Then factor it, and shortly arrive at what you want to prove.

 

[duffman868]

yes that is what i mean i will try that

 

[d_leet]

that was the way i was taught.
sec= secant=1/u
tan= tangent=u/v

so tan-4 would be (u/v)- 2

That substitution is one of two things, either completely wrong, or very unusual, put everything in terms of sine and cosine, that might help better, and I still have no clue what "tan-4" means.

Because if secant is 1/u, thangent cannot be u/v it would be v/u assuming that v = sine and u = cosine.

 

[VietDao29]

that was the way i was taught.
sec= secant=1/u
tan= tangent=u/v

so tan-4 would be (u/v)- 2

Nooooo, you again forget the angle, the tangent, secant values of what angle? (x, y, t, k, $\alpha, \ \beta, \ \gamma, \ \theta$ or what?)
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EDIT: Have you solved the problem?

 

[duffman868]

i relised that after i posted it is v/u also that is what i used in the first problem. and no i have not solved it yet i think not taking algebra 1 or 2 is hurting me now

 

[duffman868]

ok i don't know if you can do this if so why did my answer come out flipped
1+ tan^2(x) -6 tan(x) +7/ 1+ tan^2(x) -5 to
tan^2(x) -5 tan(x) +7/ tan^2(x) -4 to
tan(x) -5 tan(x) +7/ tan(x) -4 to
tan(x) +2/ tan (x) -4

 

[duffman868]

1+ tan^2(x) -6 tan(x) +7/ 1+ tan^2(x) -5 this is what i get after changing them to tangents
i remember doing quadratic equation once but don't remember what they were and i thought i could factor but can't see it in this problem. got any links handy that can help with this problem

edit there was a post above mine
i think that by factoring you want me to see that 1 + tan^2(x) can be factored like a^2+b^2 but i can't remember if there are others that apply to this problem

 

[d_leet]

1+ tan^2(x) -6 tan(x) +7/ 1+ tan^2(x) -5 this is what i get after changing them to tangents
i remember doing quadratic equation once but don't remember what they were and i thought i could factor but can't see it in this problem. got any links handy that can help with this problem

edit there was a post above mine
i think that by factoring you want me to see that 1 + tan^2(x) can be factored like a^2+b^2 but i can't remember if there are others that apply to this problem

Combine like terms and you should find that those expressions factor very nicely.

 

[VietDao29]

1+ tan^2(x) -6 tan(x) +7/ 1+ tan^2(x) -5 this is what i get after changing them to tangents
i remember doing quadratic equation once but don't remember what they were and i thought i could factor but can't see it in this problem. got any links handy that can help with this problem

edit there was a post above mine

i relised that after i posted it is v/u also that is what i used in the first problem. and no i have not solved it yet i think not taking algebra 1 or 2 is hurting me now

Arrghh, you should take Algebra before taking pre-calculus! i.e, you should have some solid, and basic knowledge of algebra, or you'll not learn much in pre-calculus. You can either find some Algebra book and study it by yourself, or you can ask a tutor... (a tutor maybe better).
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You know what addition and multiplication are, right?
Commutative property of addition: a + b = b + a, ie two numbers add to the same thing whichever order you add them in.
Example:
2 + 3 = 5
3 + 2 = 5
Commutative property of multiplication: ab = ba
3 . 5 = 15
5 . 3 = 15
Associative property of addition: (a + b) + c = a + (b + c), ie, if you take a + b first, then add c, it's the same as you add b + c together first, and then add a.
Example:
(1 + 9) + 7 = 10 + 7 = 17
1 + (9 + 7) = 1 + 16 = 17.
Associative property of multiplication: (ab)c = a(bc).
Example:
(2 . 3) . 5 = 6 . 5 = 30
2 . (3 . 5) = 2 . 15 = 30
Distributive property of multiplication with respect to addition: a(b + c) = ab + ac
Example:
2 . (3 + 4) = 2 . 7 = 14
2 . 3 + 2 . 4 = 6 + 8 = 14
Additive inverse, or opposite, of a number n is the number which, when added to n, yields zero. We denote it to be: -n.
That means n + (-n) = 0
Example:
The additive inverse of 7 is −7, because 7 + (−7) = 0
The additive inverse of a is −a (definition).
The additive inverse of (ab) = -(ab).
Subtraction is the reverse of addition, to subtract b from a, we do as follow:
a - b = a + (-b), ie, we add a and the opposite number of b.
-----------------------
Say, you want to simplify the expression:
3x - 7y + 12x - 4y - 9y = 3x + (-7y) + 12x + (-4y) + (-9y)
= 3x + 12x + (-4y) + (-9y) + (-7y) (Associative property of addition)
= (3 + 12)x + ((-4) + (-9) + (-7))y = 15x + (-20)y = 15x - 20y.
Do the same, can you go from
$$\frac{\tan ^ 2 (x) + 1 - 6 \tan (x) + 7}{\sec ^ 2 (x) - 5}$$ to $$\frac{\tan ^ 2 (x) - 6 \tan (x) + 8}{\tan ^ 2 (x) - 4}$$?
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QUADRATIC EQUATION:
Look here for quadratic equations.
-----------------------
FACTORIZATION:
See here for some basic stuff about factorization.
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Remember that, once we've factor both numerator and denominator, we can cancel out something that both numerator, and denominator have.
Example:
$$\frac{(x - 2) (3x + 5) (x + 7)}{(6x - 5) (x - 2) (x + 7)}$$
(x - 2) and (x + 7) are in both numerator, and denominator, cancelling them out, we have:
$$\frac{(x - 2) (3x + 5) (x + 7)}{(6x - 5) (x - 2) (x + 7)} = \frac{3x + 5}{6x - 5}$$
Can you do the problem now?
My last advice is: Go buying some book, and study Algebra, or hire some good tutor.
GET ALGEBRA BOOK!

 

[duffman868]

ok i understand this what i don't is how tan(x)+2/tan(x)-4 = tan(x)-4/tan(x)+2

 

[VietDao29]

No, they are not equal:
$$\frac{\tan x + 2}{\tan x - 4} \neq \frac{\tan x - 4}{\tan x + 2}$$
Since, it's quite a long time, and you may not have worked out the problem, I think I can post the solution.
Prove that:
$$\frac{\sec ^ 2 (x) - 6 \tan (x) + 7}{\sec ^ 2 (x) - 5} = \frac{\tan (x) - 4}{\tan (x) + 2}$$
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Proof:
We will now prove that the LHS of the equation is equal to the RHS. So we will start from the LHS:
$$\frac{\sec ^ 2 (x) - 6 \tan (x) + 7}{\sec ^ 2 (x) - 5} = \frac{\tan ^ 2 x + 1 - 6 \tan x + 7}{\tan ^ 2 x + 1 - 5} = \frac{\tan ^ 2 x - 6 \tan x + 8}{\tan ^ 2 x - 4}$$
We then factor both numerator and denominator:
$$\frac{(\tan x - 2) (\tan x - 4)}{(\tan x - 2) (\tan x + 2)} = \frac{\tan x - 4}{\tan x + 2}$$ (Q.E.D)
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Can you understand this proof? And have you got an Algebra book yet? :)

 

[ovoleg]

Can you understand this proof? And have you got an Algebra book yet? :)

Oh man

We all have trouble sometimes :)

I have been working on a long homework assignment for the past 6 hours, I wish I had never taken thermodynamics