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

  • Thread starter mcs117
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In summary, you are trying to solve for x and in the process are finding the cosine and sine of x. You also need to find the two corresponding cubes.
  • #1
mcs117
3
0
Can anyone help me solve this equation,

x=theta

cot(x)--------------tan(x)
------------ + ----------------- = 1 + tan(x) + cot(x)
1- tan(x)----------1- cot(x)
 
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  • #2
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
prove this.
work:
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)

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)
 
Last edited:
  • #4
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
Hmm...
[tex]
\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} [/tex]
 
  • #6
bomba923 said:
Hmm...
[tex]
\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} [/tex]

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

-Dan
 
  • #7
mcs117 said:
prove this.
Code:
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.
 

1. What is the meaning of the equation "cot(x) / tan(x) + 1 = tan(x) + cot(x)"?

The equation represents a trigonometric identity that can be used to solve for the value of x, where x is an angle measured in radians.

2. How do you solve for x in the equation "cot(x) / tan(x) + 1 = tan(x) + cot(x)"?

To solve for x, you can use algebraic manipulation and trigonometric identities to simplify the equation and isolate x on one side. The final solution will depend on the given values for cot(x) and tan(x).

3. What are the possible solutions for x in the equation "cot(x) / tan(x) + 1 = tan(x) + cot(x)"?

There are infinitely many possible solutions for x, as it is a periodic function. However, the most commonly used solutions are in the interval [0, 2π).

4. Are there any common mistakes to avoid when solving "cot(x) / tan(x) + 1 = tan(x) + cot(x)"?

One common mistake is to forget to use the Pythagorean identity (sin²(x) + cos²(x) = 1) when simplifying the equation. It is also important to check for extraneous solutions, as some values of x may result in undefined expressions.

5. How can the equation "cot(x) / tan(x) + 1 = tan(x) + cot(x)" be applied in real-world situations?

The equation can be used in various fields such as engineering, physics, and astronomy to solve for unknown angles in trigonometric problems. It can also be used in navigation to determine the direction and distance between two points.

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