A different approach:
Consider the function $$f(z) = \frac{e^{i \alpha \tan z}}{z^{2}+\beta^{2}} , \quad (\alpha, \beta >0), $$ which has simple poles at $z= \pm i \beta$ and essential singularities at the half-integer multiples of $\pi$.
The magnitude of $e^{i \alpha \tan z}$ is $$\exp \left(\frac{-\alpha \sinh\left(2 \Im (z)\right)}{\cos\left(2 \Re (z)\right)+ \cosh \left(2 \Im (z)\right)} \right). $$
In the upper half plane, this value never exceeds $1$, which includes near the essential singularities on the real axis.
So by integrating $f(z)$ around a contour consisting of the real axis (with vanishingly small indentations about the half-integer multiples of $\pi$) and the large semicircle above it, we can conclude that $$\operatorname{PV}\int_{-\infty}^{\infty} \frac{e^{i \alpha \tan x}}{x^{2}+ \beta^{2}} \, \mathrm dx = 2 \pi i \operatorname*{Res}_{z=i \beta} f(z) = \frac{\pi}{\beta } \, e^{- \alpha \tanh \beta}. $$
Equating the real parts on both sides of the above equation, we have $$\int_{-\infty}^{\infty} \frac{\cos(\alpha \tan x)}{x^{2}+\beta^{2}} \, \mathrm dx = \frac{\pi}{\beta } \, e^{- \alpha \tanh \beta}.$$
Therefore,
$$ \begin{align} \int_{0}^{\infty} \frac{\sin^{2}(\alpha \tan x)}{x^{2}} \, \mathrm dx &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin^{2}(\alpha \tan x)}{x^{2}} \, \mathrm dx \\ &= \frac{1}{2} \lim_{\beta \to 0^{+}} \int_{-\infty}^{\infty} \frac{\sin^{2}(\alpha \tan x)}{x^{2}+\beta^{2}} \, \mathrm dx \\ &= \frac{1}{4} \lim_{\beta \to 0^{+}} \int_{-\infty}^{\infty} \frac{1- \cos (2 \alpha \tan x)}{x^{2}+\beta^{2}} \, \mathrm dx \\ &= \frac{\pi }{4} \lim_{\beta \to 0^{+}} \frac{1- e^{- 2 \alpha \tanh \beta}}{\beta} = \\ &= \frac{\pi}{4} \lim_{\beta \to 0^{+}} \frac{2 \alpha \operatorname{sech}^{2}(\beta) \, e^{-2 \alpha \tanh \beta}}{1} \\ &= \frac{\pi \alpha}{2}. \end{align}$$