$4.58 \sqrt{\frac{\cos 2 \alpha}{\cot^{2} \alpha-\tan^{2} \alpha}} ; 90^{\circ}<\alpha<135^{\circ}$.
### 4.58 We have

$\frac{\cos 2 \alpha}{\cot^{2} \alpha - \tan^{2} \alpha} = \frac{\cos 2 \alpha \sin^{2} \alpha \cos^{2} \alpha}{(\cos^{2} \alpha - \sin^{2} \alpha)(\cos^{2} \alpha + \sin^{2} \alpha)} = \frac{1}{4} \sin^{2} 2 \alpha$.

Since by the condition $90^{\circ} < \alpha < 135^{\circ}$, then $180^{\circ} < 2 \alpha < 270^{\circ}$ and $\sin 2 \alpha < 0$. Therefore,

$\sqrt{\frac{1}{4} \sin^{2} 2 \alpha} = -\frac{1}{2} \sin 2 \alpha$