= Solution
For two distinct symmetric fibre families use reference directions $l_\pm=(\cos\theta,\pm\sin\theta,0)^T$, with $0<\theta<\pi/2$. The second sign is needed for the orthogonality conclusion: the PDF prints only one direction. If both families literally had that same direction, their images under an invertible deformation would remain parallel, so part (ii) would be false. The following derivation states the intended two-family geometry explicitly.
Write $c=\cos\theta$, $s=\sin\theta$ and take positive $\lambda,\alpha$, as required for a positive <stretch tensor>. Applying the <inextensible fibre constraint> to either family gives
$$
\frac{\alpha^2c^2+\alpha^{-2}s^2}{\lambda}=1.
$$
With $b=\alpha^2$, multiplication by $b$ yields $c^2b^2-\lambda b+s^2=0$. Its discriminant is $\lambda^2-4c^2s^2=\lambda^2-\sin^22\theta$, so the <two-family inextensible-fibre pure stretch> has
$$
\boxed{\alpha^2=\frac{\lambda\pm\sqrt{\lambda^2-\sin^22\theta}}{2\cos^2\theta}.}
$$
This also shows why the angle endpoints need separate treatment: at $\theta=0$ the admissible root is $\alpha^2=\lambda$, and at $\theta=\pi/2$ it is $\alpha^2=1/\lambda$. The nominal two families then represent the same unoriented fibre direction, and the two-branch/orthogonality assertions do not apply.
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