Solution
Let \(x=\tan \alpha , \ y=\tan \beta , \ z= \tan \gamma\) such that \(\dfrac{-\pi}{2}<\alpha , \beta , \gamma < \dfrac{+\pi}{2}\)
\[ \frac{4\sqrt{\tan^2 \alpha + 1}}{\tan \alpha} = \frac{5\sqrt{\tan^2 \beta + 1}}{\tan \beta} = \frac{6\sqrt{\tan^2 \gamma + 1}}{\tan \gamma}\]
\[ \Rightarrow \ \frac{4}{\sin \alpha}= \frac{5}{\sin \beta}=\frac{6}{\sin \gamma}\]
Again
\[ \begin{aligned} \tan \alpha \tan \beta \tan \gamma &=\tan \alpha + \tan \beta +\tan \gamma \\ \tan \alpha (\tan \beta \tan \gamma -1) &=\tan\beta + \tan \gamma \\ -\tan \alpha &= \frac{\tan\beta + \tan \gamma }{1- \tan\beta \tan \gamma } \\ -\tan \alpha &= \tan(\beta +\gamma) \\ \tan(k\pi - \alpha) &= \tan(\beta + \gamma) \\ \alpha +\beta +\gamma &=k\pi\end{aligned} \]
Taking \(k=1\) we get \(\alpha +\beta +\gamma =\pi\) which implies that there exists a triangle whose angles are \(\alpha , \ \beta , \ \gamma\) and whose sides opposite to these angles are proportional to 4,5 and 6 respectively.
Let the sides of such a triangle be \(4k, \ 5k, \ 6k\).
If \(s\) is the semiperimeter of the triangle then
\[s=\frac{15k}{2}\]
\[\tan \frac{\alpha}{2}=\sqrt{\frac{(s-4k)(s-6k)}{s(s-5k)}}=\sqrt{\frac{\dfrac{5k}{2} \times \dfrac{3k}{2}}{\dfrac{15k}{2} \times \dfrac{7k}{2}}}=\sqrt{\frac{1}{7}}\]
\[x = \tan \alpha=\frac{2\tan \dfrac{\alpha}{2}}{1-\tan^2 \dfrac{\alpha}{2}}=\dfrac{2\sqrt{\dfrac{1}{7}}}{1-\dfrac{1}{7}}=\frac{\sqrt{7}}{3}\]
Similarly, \(y=\dfrac{5\sqrt{7}}{9}\) and \(z=3\sqrt{7}\).