I'm going to do this problem in static equilibrium a little differently than the way the answer does. See attached diagram ...
$\displaystyle \sum F_x = 0$ ... force right = force left
$N = T\sin{\alpha}$
$\displaystyle \sum F_y = 0$ ... force up = forces down
$T\cos{\alpha} + f = W$
since the sphere is about to slip, the force of static friction is a maximum $\implies f = \mu \cdot N = \mu \cdot T\sin{\alpha}$
$T\cos{\alpha} + \mu \cdot T\sin{\alpha} = W$
$\displaystyle \sum \tau = 0$ ... torque CCW = torque CW
about the point where the sphere touches the wall ...
$T \cdot 2r\sin{\alpha} = W \cdot r \implies W = 2T \sin{\alpha}$
returning to the equilibrium equation in the y-direction & substituting for $W$ ...
$T\cos{\alpha} + \mu \cdot T\sin{\alpha} = 2T \sin{\alpha}$
divide every term by $T\cos{\alpha}$ ...
$1 + \mu \cdot \tan{\alpha} = 2\tan{\alpha}$
$1 = 2\tan{\alpha} - \mu \cdot \tan{\alpha}$
$1 = \tan{\alpha}(2 - \mu)$
$\dfrac{1}{2-\mu} = \tan{\alpha}$