Help Solve x = theta: cot(x) / tan(x) + 1 = tan(x) + cot(x)

[mcs117]

Can anyone help me solve this equation,

x=theta

cot(x)--------------[/color]tan(x)
------------ + ----------------- = 1 + tan(x) + cot(x)
1- tan(x)----------[/color]1- cot(x)

 

[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"?

 

[mcs117]

prove this.

work:
1/tan(x)--------------[/color]tan(x)
------------ + -----------------
1- tan(x)----------[/color]1- 1/tan(x)

cot(x)--------------[/color]tanˆ2(x)
------------ + -----------------
1- tan(x)----------[/color]tan(x) - 1

cot(x)--------------[/color]tanˆ2(x)
------------ - -----------------
1- tan(x)----------[/color]1 - tan(x)

this is where i got stuck on.
cot(x) - Tanˆ2(x)
-----------------
1 - tan(x)

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

 

[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?

 

[bomba923]

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

 

[topsquark]

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

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

-Dan

 

[0rthodontist]

prove this.

1/tan(x)           tan(x)
------------ + -----------------
1- tan(x)        1- 1/tan(x)

cot(x)            tanˆ2(x)
------------ + -----------------
1- tan(x)         tan(x) - 1

cot(x)            tanˆ2(x)
------------ - -----------------
1- tan(x)          1 - tan(x)

cot(x) - Tanˆ2(x)
-----------------
   1 - tan(x)

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.