suppose G is a connected graph with n vertices and n-1 edges.
prove that G has at least 2 vertices with deg(v)=1 .
Use the handshaking lemma.
Recall that $\displaystyle |\mathcal{V}|$ is the number of vertices and $\displaystyle |\mathcal{E}|$ is the number of edges.
Then we know that $\displaystyle \frac{2|\mathcal{E}|}{|\mathcal{V}|}\ge m$ where $\displaystyle m$ is the minimum vertex degree.
You can use that show that if $\displaystyle |\mathcal{V}|=n-1$ then there must be at one vertex of degree one.
You are working with a connected graph. So all vertices are of degree alt least one.
You should know that $\displaystyle \sum\limits_{v \in\mathcal{V}} {\deg (v)} = 2\left| \mathcal{E} \right|$.
Now you show some effort.