We have $\frac{sin n\theta cos\theta -cos n\theta sin \theta}{sin\theta cos\theta} =n-1$ Which is $\frac{sin((n-1)\theta)}{sin\theta cos\theta}=n-1$ So $sin((n-1)\theta )=\frac{n-1}{2}sin 2\theta$ Let $\theta =\frac{\pi}{4}$ We get $sin ((n-1)\frac{\pi}{4})=\frac{n-1}{2}$ So $\frac{n-1}{2} \in \{ 1,\frac{\sqrt{2}}{2},0,-\frac{\sqrt{2}}{2},-1 \}$ Give $n=1$ or $3$