Inverse trigonometric equation

utkarshakash

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

[itex]cot^{-1} x - cot^{-1} (n^2 - x+1)=cot^-1(n-1)[/itex]

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

[itex](n^2+1-2x)(n-1) = x(n^2-x+1)[/itex]

MrWarlock616

Are you asking how to solve that equation?

SammyS

I get nearly the same thing:

[itex](n^2+1-2x)(n-1) = 1+x(n^2-x+1)[/itex]

utkarshakash

Yes I have to find the value of x

utkarshakash

But what should be the next step?

SammyS

Multiply each side out.

Get everything to the left side.

You have a quadratic equation in x.

utkarshakash

But its discriminant is way too big!

SammyS

Too big in what sense ?

What do you get for the discriminant ?

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=[itex]n^4+6n^2-8n+2[/itex]

SammyS

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

I got it because,

[itex]\displaystyle \left(1+\frac{1}{x(n^2-x+1)}\right)x(n^2-x+1)=x(n^2-x+1)+1\ .[/itex]

How did you not get it?

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

[itex] x= \dfrac{n(n+1)}{2} [/itex]

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

SammyS

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

SammyS

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

[itex]\displaystyle \cot(A+B)=\frac{\cot(A)\,\cot(B)-1}{\cot(A)+\cot(B)}[/itex]

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

[itex] -4n^3+6n^2-8n+9[/itex]

SammyS

What I got for a discriminant was

[itex]n^4+6n^2-8n+9\ .[/itex]

I just rechecked it using WolframAlpha on

[itex](n^2+1-2x)(n-1) = 1+x(n^2-x+1)[/itex]

and it confirmed my result.

utkarshakash

OK so what should I do after getting discriminant?

haruspex

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