# Math Problem

1. Mar 2, 2006

### mcs117

Can anyone help me solve this equation,

2. Mar 2, 2006

### assyrian_77

What have you done so far? Solutions are not just given away in this forum.

Is it really "solve for x" or is it "prove this"?

3. Mar 2, 2006

### mcs117

prove this.
I can't seem to get rid of the bottom and make it 1 + tan(x) + cot(x)

Last edited: Mar 2, 2006
4. Mar 2, 2006

### konartist

Hmmm, I haven't tried this problem, but just a tip, instead of writing cot as 1/tanx write it as cos/sin. Makes things simpler.

The only thing I can suggest is get common denominators on the left side.

Wait, are you trying to prove the identity?

5. Mar 2, 2006

### bomba923

Hmm...
$$\begin{gathered} \frac{{\cot x}} {{1 - \tan x}} + \frac{{\tan x}} {{1 - \cot x}} = 1 + \tan x + \cot x \Rightarrow \hfill \\ \frac{1} {{\left( {1 - \tan x} \right)\tan x}} + \frac{{\tan ^2 x}} {{\tan x - 1}} = \frac{{\tan ^2 x + \tan x + 1}} {{\tan x}} \Rightarrow \hfill \\ \frac{{1 - \tan ^3 x}} {{1 - \tan x}} = \frac{{\tan ^2 x + \tan x + 1}} {{\tan x}} \Rightarrow \hfill \\ \left( {1 - \tan ^3 x} \right)\tan x = \left( {1 - \tan x} \right)\left( {\tan ^2 x + \tan x + 1} \right) \Rightarrow \hfill \\ \tan x - \tan ^4 x = \tan ^2 x + \tan x + 1 - \tan ^3 x - \tan ^2 x - \tan x \Rightarrow \hfill \\ \tan x\left( {1 - \tan ^3 x} \right) = 1 - \tan ^3 x \Rightarrow \tan x = 1 \Rightarrow {\text{Mistake?}} \hfill \\ \end{gathered}$$

6. Mar 3, 2006

### topsquark

Line 3 should be
$$\frac{1-tan^3x}{(1-tanx)tanx}=\frac{1+tanx+tan^2x}{tanx}$$

-Dan

7. Mar 3, 2006

### 0rthodontist

Well, multiply through by tan x
(1 - tan^3(x))/((tan x)(1 - tan x))
Now you can factor the numerator as a difference of cubes, and reduce.