Flipping H/A Fraction for Canceling Out A's

[supernova1203]

o/a + h/a =1

how do i flip the h/a fraction? Do i divide them instead of adding them? (Im trying to cancel out the 2 a values)

 

[mharten1]

I'm not sure what you're asking...why do you want to cancel out the a values? And what are you supposed to do with that equation? Simplify, solve for a variable, etc?

 

[supernova1203]

I'm not sure what you're asking...why do you want to cancel out the a values? And what are you supposed to do with that equation? Simplify, solve for a variable, etc?

Identities for trig ratios

 

[mharten1]

Identities for trig ratios

Perhaps I'm just misreading what you initially wrote, but wouldn't o/a + h/a = 1 be the same as tan(θ) + sec(θ) = 1 ? That's not an identity...

 

[Mark44]

o/a + h/a =1

how do i flip the h/a fraction? Do i divide them instead of adding them? (Im trying to cancel out the 2 a values)

It's not clear what you're trying to do when you ask how to "flip fractions."

o/a + h/a = 1
<==> (o + h)/a = 1
<==> a/(o + h) = 1/1 = 1 as long as o + h != 0

You can solve for a, if that's what you're trying to do, by multiplying both sides of the equation by o + h.

a = 1* (o + h) = o + h

One thing you CANNOT DO is just "flip" the fractions. For example,
1/2 + 1/2 = 1
but 2/1 + 2/1 is not equal to 1/1.

 

[Dembadon]

o/a + h/a =1

how do i flip the h/a fraction? Do i divide them instead of adding them? (Im trying to cancel out the 2 a values)

Can you post the original problem using the template?

You cannot flip h/a in this particular equation. Yes, dividing by h/a is the same as multiplying by a/h. However, without seeing your problem, we have no way of knowing whether or not you'd be murdering the equation if you were to change it.

 

[HallsofIvy]

Apparently "o", "a", and "h" are "opposite side", "adjacent side" and "hypotenuse", respectively in a right triangle. It would have been nice to tell us that.

What you can do is first solve for "h/a":
\frac{h}{a}= 1- \frac{o}{a}= \frac{a- o}{a}
and the "flip" both sides:
\frac{a}{h}= \frac{a}{a- o}
but that right side is not any trig function.

As mharten1 said, if these really are trig ratios, then what you have is
cot(\theta)+ csc(\theta)= 1
but that is NOT, in general, true!

 

[Mentallic]

As mharten1 said, if these really are trig ratios, then what you have is
cot(\theta)+ csc(\theta)= 1
but that is NOT, in general, true!

And it is true only in the degenerate case where h=a-o, thus where the triangle collapses into a line.

 

[zketrouble]

o/a + h/a =1

how do i flip the h/a fraction? Do i divide them instead of adding them? (Im trying to cancel out the 2 a values)

If you are trying to cancel out the a's, I would multiply both sides of the equation by a.

o/a + h/a = 1
(o+h)/a = 1
o+h = 1a
o+h = a

And if you are using this for a geometery problem, I think there might be an error somewhere. Opposite + Hypotenuse = Adjacent? shouldn't it be B^2 = C^2 - A^2?? Please forgive if this I'm wrong about what you were intending to apply this to.