# calculation of $x$ in $\displaystyle \tan^{-1}\left(\frac{x}{1-x^2}\right)+\tan^{-1}\left(\frac{1}{x^3}\right) = \frac{3\pi}{4}$

The no. of real values of $x$ satisfying $\displaystyle \tan^{-1}\left(\frac{x}{1-x^2}\right)+\tan^{-1}\left(\frac{1}{x^3}\right) = \frac{3\pi}{4}$

options ::

(a) $0$

(b) $1$

(c) $2$

(d) Infinitely many

My Try:: Using The formula $\tan^{-1}A+\tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$

So $\displaystyle \tan^{-1}\left(\frac{\frac{x}{1-x^2}+\frac{1}{x^3}}{1-\frac{x}{x^3.(1-x^2)}}\right) = \frac{3\pi}{4}$

So $\displaystyle \frac{x^4+1-x^2}{x^3-x^5-x}= -1$

So $(x-1)(x^4-x+1) = 0$

$x = 1$ and $x^4 -x+1 = 0$

So Iam getting only one value of $x$ which is $x=1$ and How can i calculate any real value of $x$ exists from $x^4-x+1 = 0$

OR any other method by using we can solve this Question

Thanks

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The equation $x^4-x+1=0$ has no real solution. – Hanul Jeon Jan 26 '13 at 10:28
@juantheron I noticed you've asked several question on MSE, but you've never accepted an answer. You should accept your favorite answer if it is satisfactory. – Git Gud Jan 26 '13 at 10:46
O sorry Git Gud .... – juantheron Jan 26 '13 at 11:15

1. If $|x|<1$, then $x^4+1>1>x$. If $|x|\geq 1$, then $x^4+1>x^4\geq x$. In both cases $$x^4-x+1>0$$ so the equation $x^4-x+1=0$ have no real solutions.

2. Note that $x=1$ is not a solution of original equation.

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Thanks Norbert for solving $x^4-x+1 = 0$ but i did not understand why $x= 1$ is not a solution of original equation because at $x = 1$ the equation is $\displaystyle \tan^{-1}(\infty)+\tan^{-1}(1) = \frac{\pi}{2}+\frac{\pi}{4} = \frac{3\pi}{4}$ which is exactly equaal to $R.H.S$ – juantheron Jan 26 '13 at 10:46
Expression $\tan^{-1}(\infty)$ makes no sense. Even if you are talking about limits it is not ok because $$\lim\limits_{x\to 1+}\tan^{-1}\frac{x}{1-x^2}=\frac{\pi}{2}$$$$\lim\limits_{x\to 1-}\tan^{-1}\frac{x}{1-x^2}=-\frac{\pi}{2}.$$ Not to mention that you can't substitute $x=1$ into expression $\frac{x}{1-x^2}$ – Norbert Jan 26 '13 at 10:51
Oh Thanks Norbert i did not consider it means $L.H.S$ and $R.H.S$ at $x=1$ So our final answer of the question is no value of $x$ means option (a) – juantheron Jan 26 '13 at 11:16