solve for $x$ in inverse trignometry

Question

$$ \operatorname{arccot} x + \operatorname{arccot} (n^2-x + 1) = \operatorname{arccot }(n - 1) $$ In this we have to solve for value of $x$ .

I thought to convert arccot into arctan . Then add using the identity .

But its getting too long .

Is there any short method to solve it.

Answer 1

1) Take the cotangent from both side:

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

2) Use $cot(A+B)={cot(A)cot(B)-1\over cotA+cotB}$

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

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

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

$A:=n^2+1$

$x^2-Ax+(A(n-1)+1)$

$x= {A\pm\sqrt{A^2-4(A(n+1)+1)}\over2}$=$ {(n^2+1)\pm\sqrt{(n^2+1)^2-4((n^2+1)(n+1)+1)}\over2}$

$x_{1}=n$
$x_{2}=n^2-n+1$

Answer 2

HINT.-Take tangent in both sides of $$\operatorname{arccot} x + \operatorname{arccot} (n^2-x + 1) = \operatorname{arccot }(n - 1)$$ so you have $$\frac{\frac1x+\frac{1}{n^2-x+1}}{1-\frac{1}{x(n^2-x+1)}}=\frac{1}{n-1}\Rightarrow x^2-(n^2+1)x+[(n-1)(n^2+1)+1]=0$$ a quadratic equation you can solve.