Determine the angle between the left and right tangents

Determine the angle between the left and right tangents of the curve $y=\arcsin\dfrac{2x}{1+x^{2}}$ at the point $x=1$.

Step 1: Unfold the composite function

With $x=\tan\theta$, $\theta\in\left(-\tfrac\pi2,\tfrac\pi2\right)$, the double-angle identity gives $\dfrac{2x}{1+x^{2}}=\dfrac{2\tan\theta}{1+\tan^{2}\theta}=\sin2\theta$. Since $\arcsin(\sin2\theta)=2\theta$ only while $2\theta\in\left[-\tfrac\pi2,\tfrac\pi2\right]$, i.e.\ $|x|\le1$, the function is genuinely piecewise:

\[ y = \begin{cases} 2\arctan x, & |x|\le1,\\[4pt] \pi-2\arctan x, & x\gt 1,\\[4pt] -\pi-2\arctan x, & x\lt -1. \end{cases} \tag{1} \]

(The expression $\dfrac{2x}{1+x^2}$ itself always lies in $[-1,1]$ and is smooth; it is the outer $\arcsin$ that forces a corner at $x=\pm1$, where the argument reaches $\pm1$.)

Step 2: Derivatives on each side of $x=1$

For $|x|\lt 1$: $y=2\arctan x \Rightarrow y'=\dfrac{2}{1+x^{2}}$. For $x\gt 1$: $y=\pi-2\arctan x \Rightarrow y'=-\dfrac{2}{1+x^{2}}$.

Step 3: One-sided slopes at $x=1$

\[ k_{-} = \lim_{x\to1^{-}}\frac{2}{1+x^{2}} = \frac{2}{2}=1, \qquad k_{+} = \lim_{x\to1^{+}}\left(-\frac{2}{1+x^{2}}\right) = -\frac{2}{2}=-1. \tag{2} \]

The function has a corner at $x=1$, with left slope $1$ and right slope $-1$.

Step 4: Angle between the two tangents

\[ \tan\alpha = \left|\frac{k_{+}-k_{-}}{1+k_{-}k_{+}}\right| = \left|\frac{-1-1}{1+(1)(-1)}\right| = \left|\frac{-2}{0}\right| \to\infty. \tag{3} \]

The denominator vanishes, so $\alpha=\dfrac\pi2$: the two directions $(1,1)$ and $(1,-1)$ have dot product $1-1=0$, confirming perpendicularity directly.

Step 5: Verification

\[x=0.999\text{: } y\approx1.569796,\ 2\arctan(0.999)\approx1.569796\ \checkmark\]

\[x=1.001\text{: } y\approx1.569797,\ \pi-2\arctan(1.001)\approx1.569797\ \checkmark\]

\[\text{One-sided difference quotients at } x=1\text{: left} \approx0.99993,\ \text{right}\approx-1.00004\ \checkmark\]