If the line $y=mx+1$ is a tangent to $\frac{x^2}{100}-\frac{y^2}{49}=1$, then

$y=mx+c$ is a tangent to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ then

$c^2=a^2{m}^2-b^2$ (condition)

$c=1,a^2=100,b^2=49$

 $1^2=100m^2-49$ (tangent condition)

$\begin{array}{l}\Rightarrow 0=100m^2-50\\ \Rightarrow 0=2m^2-1\\ \Rightarrow 2m^2=1\\ \Rightarrow m^2=\frac{1}{2}\Rightarrow m=\pm \frac{1}{\sqrt{2}}\end{array}$