Помогите со вторым заданием

Решите задачу:

$\sin\alpha=\frac{24}{25},\;0\ \textless \ \alpha\ \textless \ \frac\pi2\\\sin^2\alpha+\cos^2\alpha=1\Rightarrow\cos\alpha=\sqrt{1-\sin^2\alpha}\\\cos^2\alpha=\sqrt{1-\frac{576}{625}}=\sqrt{\frac{49}{625}}=\pm\frac7{25}\\0\ \textless \ \alpha\ \textless \ \frac\pi2\Rightarrow\cos\alpha\ \textgreater \ 0\\\cos\alpha=\frac7{25}\\\sin2\alpha=2\sin\alpha\cos\alpha=2\cdot\frac{24}{25}\cdot\frac7{25}=\frac{336}{625}\\ctg2\alpha=\frac{\cos2\alpha}{\sin2\alpha}\\\cos2\alpha=\cos^2\alpha-\sin^2\alpha=\frac{49}{625}-\frac{576}{625}=-\frac{527}{625}$
$ctg2\alpha=-\frac{527}{625}:\frac{24}{25}=-\frac{527}{625}\cdot\frac{25}{24}=-\frac{527}{25}\frac1{24}=-\frac{527}{600}$