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Consider a matrix $B\in \mathrm{SL}(2,\mathbb R)$. Let $s$ be a vector that is pointing in the most contracted direction of $B$, and let $u$ be the image under $B$ of a unit vector pointing in the most expanded direction of $B$. (The lengths of the vectors $s$ and $u$ don't really matter)

Let $z$ be a nonzero vector not pointing in the most contracted direction. For some large positive parameter $r$ let $\theta\in (r^{-80},\frac{\pi}{2}]$ be the angle between $z$ and $s$, and let $\varphi\in [0,\frac{\pi}{2}]$ be the angle between $Bz$ and $u$.

Let $\Vert B\Vert$ denote the norm of the largest eigenvalue of $B$. Assume that $\Vert B\Vert\geq e^{cr}$ for some constant $c$ and $r$ sufficiently large. How do we show that $\varphi<r^{-100}$?


Context: this is a step to the proof of Claim 3.7 in this paper (if you have access to Comm Math Phys, the published version is here). In that paper, they used the "identity" $ \tan (\theta)\tan(\varphi)=\frac{1}{\Vert B \Vert^2},$ which I am pretty sure is false (you can choose $z$ to be in the most expanding direction, in which case we have $\varphi=0$ and so the LHS is $0$).

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