simplifying a complex fraction

**cocoknny** · October 17th 2007, 03:50 PM

h=(Tanά)d x tanβx
Tanά-tanβ

**topsquark** · October 17th 2007, 04:19 PM

Originally Posted by cocoknny

h=(Tanά)d x tanβx
Tanά-tanβ

Is this
$h = \frac{tan(\alpha)}{tan(\alpha) - tan(\beta)} \times tan(\beta x)$

If not your expression is too unclear to work with. Could you possibly write it out and scan it?

-Dan

**cocoknny** · October 17th 2007, 05:59 PM

hello, um what u posted is right

um i didn't have the functions u have on ur computer to do mine like yours. If u can assist in solving that fraction in anyway it is greatly appreciated.

**cocoknny** · October 17th 2007, 06:00 PM

in this one i have to change tan to cotangent.

**angel.white** · October 17th 2007, 08:23 PM

Originally Posted by cocoknny

in this one i have to change tan to cotangent.

I can't tell what your equation is, but I hope this formula will help

This is the subtraction formula for tangent
$tan(x-y) = \frac{tan(x) - tan(y)}{1+tan(x)tan(y)}$

If yours is the cotangent, it should be
$\frac{1}{tan(x-y)} = \frac{1}{\frac{tan(x) - tan(y)}{1+tan(x)tan(y)}}$

or
$cot(x-y) = \frac{1+tan(x)tan(y)}{tan(x) - tan(y)}$

because $cot(x) = \frac{1}{tan(x)}$

**Jhevon** · October 17th 2007, 08:25 PM

Originally Posted by cocoknny

hello, um what u posted is right

...

if so, what was thar "d" after $\tan \alpha$ that you had in your problem that topsquark did not write in his? please double check to make sure you have the correct thing, we don't want to tackle a problem only to find out that we solved the wrong one

**Soroban** · October 18th 2007, 07:28 AM

Hello, cocoknny!

If that's really $\beta x$ , we've got trouble ... with a capital T.

Simplify:. $h\;=\;\frac{\tan\alpha\tan\beta}{\tan\alpha - \tan\beta}$

It doesn't really simplify . . . we can only exchange it for another expression.

We have: . $\Large{\frac{\frac{\sin\alpha}{\cos\alpha}\cdot\fr ac{\sin\beta}{\cos\beta}}{\frac{\sin\alpha}{\cos\a lpha} - \frac{\sin\beta}{\cos\beta}}}$

Multiply top and bottom by $\cos\alpha\cos\beta\!:\;\;\frac{\sin\alpha\sin\bet a}{\sin\alpha\cos\beta - \sin\beta\cos\alpha}$

. . And we have: . $\frac{\sin\alpha\sin\beta}{\sin(\alpha - \beta)}$

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