11.50 What is the value of the fraction $\frac{1-\sin \alpha}{\cos \alpha}$, if $\operatorname{ctg} \frac{\alpha}{2}=m$?
11.50 Denoting the given expression by $A$, we will transition to the argument $\frac{\alpha}{2}$ and perform the following transformations:

$$
\begin{aligned}
& A=\frac{\cos ^{2} \frac{\alpha}{2}+\sin ^{2} \frac{\alpha}{2}-2 \cos \frac{\alpha}{2} \sin \frac{\alpha}{2}}{\cos ^{2} \frac{\alpha}{2}-\sin ^{2} \frac{\alpha}{2}}= \\
& =\frac{\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)^{2}}{\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)\left(\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}\right)}
\end{aligned}
$$

The expression $A$ is defined under the condition

$\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)\left(\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}\right) \neq 0$.

Solving the equations $\cos \frac{\alpha}{2} \pm \sin \frac{\alpha}{2}=0$. Since $\operatorname{ctg} \frac{\alpha}{2}=m$ exists, then $\sin \frac{\alpha}{2} \neq 0$. Dividing both sides of the equations by $\sin \frac{\alpha}{2}$, we get $\operatorname{ctg} \frac{\alpha}{2} \pm 1=0$, from which $\operatorname{ctg} \frac{\alpha}{2}= \pm 1=m$. Therefore, the expression $A$ is defined under the condition $m \neq \pm 1$. Next, by canceling the fraction (1) by $\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}$ and dividing the numerator and the denominator by $\sin \frac{\alpha}{2} \neq 0$, we find

$$
A=\frac{\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}}=\frac{\operatorname{ctg} \frac{\alpha}{2}-1}{\operatorname{ctg} \frac{\alpha}{2}+1}=\frac{m-1}{m+1}
$$

Answer: $\frac{m-1}{m+1}$.