Inverse trigonometric equation

[utkarshakash]

Homework Statement

Solve

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

Homework Equations

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

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

 

[MrWarlock616]

Are you asking how to solve that equation?

 

[SammyS]

Homework Statement

Solve

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

Homework Equations

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

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

I get nearly the same thing:

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

 

[utkarshakash]

Are you asking how to solve that equation?

Yes I have to find the value of x

 

[utkarshakash]

I get nearly the same thing:

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

But what should be the next step?

 

[SammyS]

But what should be the next step?

Multiply each side out.

Get everything to the left side.

You have a quadratic equation in x.

 

[utkarshakash]

Multiply each side out.

Get everything to the left side.

You have a quadratic equation in x.

But its discriminant is way too big!

 

[SammyS]

But its discriminant is way too big!

Too big in what sense ?

What do you get for the discriminant ?

 

[utkarshakash]

Too big in what sense ?

What do you get for the discriminant ?

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

 

[SammyS]

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'.
...]

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?

 

[utkarshakash]

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?

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

 

[SammyS]

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

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

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.

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

 

[utkarshakash]

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

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

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

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

 

[SammyS]

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

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

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.

 

[utkarshakash]

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.

OK so what should I do after getting discriminant?

 

[haruspex]

OK so what should I do after getting discriminant?

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

 

[utkarshakash]

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

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?

 

[SammyS]

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?

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.

 

[utkarshakash]

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.

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

 

[SammyS]

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

That's about all there is to do, but you can't do much simplifying.

 

[utkarshakash]

That's about all there is to do, but you can't do much simplifying.

OK Thanks!