# Inverse trigonometric equation

1. Jan 5, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
Solve

$cot^{-1} x - cot^{-1} (n^2 - x+1)=cot^-1(n-1)$

2. Relevant equations

3. The attempt at a solution
I can write the above equation as θ+α=β where the symbols represent the respective inverse functions. Now I take tan of both sides. Simplifying I get

$(n^2+1-2x)(n-1) = x(n^2-x+1)$

2. Jan 5, 2013

### MrWarlock616

Are you asking how to solve that equation?

3. Jan 5, 2013

### SammyS

Staff Emeritus
I get nearly the same thing:

$(n^2+1-2x)(n-1) = 1+x(n^2-x+1)$

4. Jan 6, 2013

### utkarshakash

Yes I have to find the value of x

5. Jan 6, 2013

### utkarshakash

But what should be the next step?

6. Jan 6, 2013

### SammyS

Staff Emeritus
Multiply each side out.

Get everything to the left side.

You have a quadratic equation in x.

7. Jan 6, 2013

### utkarshakash

But its discriminant is way too big!

8. Jan 6, 2013

### SammyS

Staff Emeritus
Too big in what sense ?

What do you get for the discriminant ?

9. Jan 7, 2013

### utkarshakash

How are you getting that 'extra' 1 in the RHS. OK I'm only posting what I got from my answer i.e. without that '1'.

D=$n^4+6n^2-8n+2$

10. Jan 7, 2013

### SammyS

Staff Emeritus
Wow! I showed that "extra" 1 way back in post #3. Now you mention it?

I got it because,

$\displaystyle \left(1+\frac{1}{x(n^2-x+1)}\right)x(n^2-x+1)=x(n^2-x+1)+1\ .$

How did you not get it?

11. Jan 8, 2013

### utkarshakash

Hey there is a much better solution than this. Instead of taking tan if I take cot of both sides I get a very simple result and here's it

$x= \dfrac{n(n+1)}{2}$

And yes you said right. There was a calculation error in my solution. It should have that extra 1

12. Jan 8, 2013

### SammyS

Staff Emeritus
That is interesting. I'll have to take a look at it.

13. Jan 9, 2013

### SammyS

Staff Emeritus
I still get a quadratic equation in x, so that can't be the solution.

$\displaystyle \cot(A+B)=\frac{\cot(A)\,\cot(B)-1}{\cot(A)+\cot(B)}$

14. Jan 9, 2013

### utkarshakash

OMG I made the silliest mistake ever. Instead of taking cot of both sides I just removed the 'inverse thing' and simply calculated the value of x from there. OK sticking to my previous solution I take tan of both sides and get the same thing what you got(with that extra 1) and get the discriminant as

$-4n^3+6n^2-8n+9$

15. Jan 9, 2013

### SammyS

Staff Emeritus
What I got for a discriminant was

$n^4+6n^2-8n+9\ .$

I just rechecked it using WolframAlpha on

$(n^2+1-2x)(n-1) = 1+x(n^2-x+1)$

and it confirmed my result.

16. Jan 9, 2013

### utkarshakash

OK so what should I do after getting discriminant?

17. Jan 9, 2013

### haruspex

Nothing special. It's just a quadratic equation in x; solve it with the usual formula. If it's messy, it's messy.

18. Jan 10, 2013

### utkarshakash

You can't simply say that I cannot get rid of that square root. Please help me. Its not even a perfect square. Should I leave it this way?

19. Jan 10, 2013

### SammyS

Staff Emeritus
If n=0, then the discriminant is a perfect square. I found no other cases. Also, the discriminant is positive for all values of n.

So as haruspex said, it's a messy result.

20. Jan 10, 2013

### utkarshakash

OK so you want me to continue with that square root.