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The equation is

$$ \sin\left(\frac{x}{x-1}\right) + 2 \tan^{-1}\left(\frac{1}{x+1}\right)=\frac{\pi}{2} $$

The answer is $0$, but I do not know how they got that.

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Is the first term $\sin^{-1} \left ( \frac{x}{x-1} \right )$? – Ron Gordon Jan 10 '13 at 17:22
Right-click on any MathJax formula, select Show Math As and TeX Commands, and you’ll see how the formula was coded. – Brian M. Scott Jan 10 '13 at 17:23
@Brian: I know, but given the problem and the expectations of the solution, I sense a disconnect. – Ron Gordon Jan 10 '13 at 17:34
@rlgordonma: That comment was for the OP, who mentioned not knowing how to get the exponent to display properly. – Brian M. Scott Jan 10 '13 at 17:38


as $$2\tan^{-1}y=\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)$$ (Proof below)

$$\implies \sin^{-1}\left(\frac x{x-1}\right)=\frac\pi2-\cos^{-1}\left(\frac{x^2+2x}{x^2+2x+2}\right)=\sin^{-1}\left(\frac{x^2+2x}{x^2+2x+2}\right)$$

So, $$\frac x{x-1}=\frac{x^2+2x}{x^2+2x+2}$$

Now, solve for $x$

[Proof: $$\tan^{-1}y=z\implies y=\tan z, \cos 2z=\frac{1-\tan^2z}{1+\tan^2z}=\frac{1-y^2}{1+y^2}\implies 2z=\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)\implies 2\tan^{-1}y=\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)$$]

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You've got a very nice fact in the end of the answer. I didn't see like that. Thanks soooooo much for sharing us that. +1 – Babak S. Jan 10 '13 at 17:49
#neverloggedin, $x=-4$ is also another solution – lab bhattacharjee Jan 10 '13 at 17:57

For a numerical aproximation, set $F(x)=\sin\left(\frac{x}{x-1}\right) + 2 \tan^{-1}\left(\frac{1}{x+1}\right)-\pi/2$. Use the Newton-Rapshon methold for give the root whit convenient start point $x_0$: $$ x_{n+1}=x_n-\frac{ F(x_n)}{F^\prime(x_n)}\quad n=0,1,2,\ldots $$

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At $x=0$ the function $F'(x)$ is really vanished. – Babak S. Jan 10 '13 at 17:56

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