Q.The angle between the curve $y=e^{mx}$ and $y=e^{-mx}$ for $m>1$ is

Both curves meet only at $(0,1)$; computing their slopes there and applying the angle-between-two-curves formula gives $\tan^{-1}\left(\dfrac{2m}{1-m^2}\right)$.

1. Intersection: solve $e^{mx}=e^{-mx}\Rightarrow e^{2mx}=1\Rightarrow 2mx=0\Rightarrow x=0$ (since $m\neq0$). At $x=0$, $y=e^{0}=1$, so the curves meet only at $(0,1)$.
2. Slope of $y=e^{mx}$: $\dfrac{dy}{dx}=me^{mx}$; at $x=0$: $m_1=m$.
3. Slope of $y=e^{-mx}$: $\dfrac{dy}{dx}=-me^{-mx}$; at $x=0$: $m_2=-m$.
4. Angle between two curves at their common point uses the angle-between-lines (tangents) formula: $\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|=\left|\frac{m-(-m)}{1+m(-m)}\right|=\left|\frac{2m}{1-m^2}\right|$ …