Let $A$ be a $2 \times 2$ matrix with $\mathrm{det}(A)=1$, two eigenvalues $\lambda_1,\,\lambda_2$ with $\lambda_1<1<\lambda_2$ and the corresponding eigenvectors to $\lambda_1,\, \lambda_2$ being the horizontal and vertical axis respectively. If $L \subset \mathbb{R}^2$ is a line with positive slope, a unit-neighbourhood of $L$ is mapped under $A$ to $C$-neighbourhood of the line $A(L)$ where $C>0$ is a not-yet-known constant.
In a proof I am reading about the properties of a map related to $A$, they give a bound on $C$ that I cannot derive myself. If $\alpha$ is the angle of $L$ to the horizontal and $\beta$ the angle of $A(L)$ to the horizontal, it is stated that \begin{equation*} C \leq \lambda_1 \frac{\sin(\beta)}{\sin(\alpha)}. \end{equation*}
My approach was to consider a unit normal vector to $L$, find its image under $A$ and to project this down onto a unit normal of $A(L)$. While I believe this gives a correct value for $C$, it is harder to deduce information from this larger expression (and so it is not as desirable for the proof) than the bound above. How can we show the above bound?
Let $L:y=mx+b$ with $\tan\alpha=m$. We can work out the unit neighbourhood of $L$ as $y=mx+b\pm\sec\alpha$.*
From the given conditions $$A=\begin{bmatrix}\lambda_1&0\\0&\lambda_1^{-1}\end{bmatrix}$$ and so $A$ maps $(x,y)$ to $(x',y')=(\lambda_1x,\lambda_1^{-1}y)$. $$\lambda_1y'=m\lambda_1^{-1}x'+b\pm\sec\alpha$$ $$y'=m\lambda_1^{-2}x'+b\lambda_1^{-1}\pm\lambda_1^{-1}\sec\alpha$$ which gives $$\tan\beta=m\lambda_1^{-2}$$ $$C=\lambda_1^{-1}\frac{\cos\beta}{\cos\alpha}^*$$ Take the quotient of the two tangents: $$\frac{\tan\beta}{\tan\alpha}=\lambda_1^{-2}=\frac{\sin\beta}{\sin\alpha}\frac{\cos\alpha}{\cos\beta}$$ $$\lambda_1^{-2}=\frac{\sin\beta}{\sin\alpha}\frac1{C\lambda_1}$$ $$C=\lambda_1\frac{\sin\beta}{\sin\alpha}$$ as desired.
*For pictorial derivations, see the below images:
