My direction is turning the problem into
$$\frac{\tan(10^\circ)}{10} \space\space\space\space?\space\space\space\space \frac{\tan(7^\circ)}{7}$$ then differentiating $f(x)=\frac{\tan(x)}{x}$, which is $$f'(x)=\frac{\frac{x}{cos^2(x)}-tan(x)}{x^2}$$
but I don't know how to continue from here.
$f'(x)=[x-\sin x \cos x]/[x^2\cos^2 x]$ And $\sin x\cos x\leq \sin x\leq x$ for $x\in (0,\frac{\pi}{2})$. So, $f'(x)>0$ in that interval, i e. $f$ is increasing in that interval. Your angles in radians are in that interval, so you can compare the two expressions that you want.
$$f'(x)=\frac{\frac{x}{\cos^2(x)}-\tan(x)}{x^2}=\frac{x-\sin x \cos x}{x^2\cos^2x}=\frac{2x-2 \sin x \cos x}{2x^2\cos^2x}=\frac{2x-\sin 2x}{2x^2\cos^2x}$$
Now to prove that $f'(x)>0$ you need to prove that $2x\gt\sin 2x$ or $\alpha>\sin\alpha$. And this is really trivial. For example, check a few answers given here: Why $x<\tan{x}$ while $0<x<\frac{\pi}{2}$?
