$\newcommand{\tr}{\operatorname{tr}}$
Let $k \subset E \subset K$ be extension fields. Show that
$$\tr\ \deg (K/k)= \tr\ \deg (K/E) + \tr\ \deg (E/k).$$
Here if we consider $S_1$ and $S_2$ be two transcedence base of $K/E $ and $E/k$ then $S_1 \cup S_2$ is so for $K/k$. My question is it trivially holds that $S_1\cup S_2$ is algebraically dependent? If not then how can I prove the algebraic dependency part?
Let $S_1 = \{x_1,\ldots,x_n\}$ and $S_2 = \{y_1\ldots y_m\}$, and suppose there is a polynomial relation $P(x_1,\ldots,x_n,y_1,\ldots,y_m) = 0$ with coefficients in $k$.
Since $k[x_1\ldots x_n,y_1 \ldots y_m] = k[y_1\ldots y_m][x_1\ldots x_n] \subset E[x_1 \ldots x_n]$, $P$ can also be viewed as a polynomial relation $P(x_1,\ldots,x_n)$ with coefficients in $E$. Since $S_1$ is a transcendance basis of $K$ over $E$, this implies that its coefficients are all zero.
Since each coefficient is a polynomial $Q(y_1,\ldots y_m)$ with coefficients in $k$ and since $S_2$ is a transcendance basis of $E$ over $k$, those coefficients can only be $0$ if all the coefficients of each $Q$ are $0$.
This shows that all the coefficients of $P$ are zero, and so there is no nontrivial polynomial relation between $S_1$ and $S_2$