3.377. Find the value of the expression $\frac{1+\cos 2 \alpha}{\operatorname{ctg} \frac{\alpha}{2}-\operatorname{tg} \frac{\alpha}{2}}$, given that $\sin \alpha+\cos \alpha=m$.
Solution.

$$
\begin{aligned}
& \frac{1+\cos 2 \alpha}{\cot \frac{\alpha}{2}-\tan \frac{\alpha}{2}}=\frac{1+\cos ^{2} \alpha-\sin ^{2} \alpha}{\frac{\cos \frac{\alpha}{2}}{\sin \frac{\alpha}{2}}-\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}}=\frac{2 \cos ^{2} \alpha}{\frac{\cos ^{2} \frac{\alpha}{2}-\sin ^{2} \frac{\alpha}{2}}{\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}}=\frac{2 \cos ^{2} \alpha}{\frac{\cos \alpha}{\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}}= \\
& =\frac{2 \cos ^{2} \alpha \cdot \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\cos \alpha}=\cos \alpha\left(2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}\right)=\cos \alpha \sin \alpha
\end{aligned}
$$

By squaring both sides of the equation $\sin \alpha+\cos \alpha=m$, we get $\sin ^{2} \alpha+2 \sin \alpha \cos \alpha+\cos ^{2} \alpha=m^{2}$, from which $\sin \alpha \cos \alpha=\frac{m^{2}-1}{2}$.

Answer: $\frac{m^{2}-1}{2}$