Let $\mathrm{S}$ and ${\mathrm{S}}^1$ are the foci and ${\mathrm{BB}}^1$ be the minor axis of an ellipse. If $\overline{)\angle {\mathrm{BSS}}^1=\theta }$ the $\mathrm{e}=$

Let ellipse be $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$

since $\angle {\mathrm{BSS}}^1=\mathrm{\theta }$ then 

slope of $\mathrm{BS}=\tan ({180}^0-\mathrm{\theta })$

$$\begin{gathered} \Rightarrow \frac{-\mathrm{b}}{\mathrm{ae}} = -\mathrm{tan\theta } \\ \Rightarrow \mathrm{b}=\mathrm{ae} \tan \mathrm{\theta } \\ \Rightarrow {\mathrm{b}}^2={\mathrm{a}}^2{\mathrm{e}}^2 {\tan }^2\mathrm{\theta } \end{gathered}$$       

          $$\begin{gathered} \overline{)\Rightarrow {\mathrm{a}}^2(1-{\mathrm{e}}^2)}={\mathrm{a}}^2{\mathrm{e}}^2{\tan }^2\mathrm{\theta } \\ \overline{) \Rightarrow 1-{\mathrm{e}}^2={\mathrm{e}}^2{\tan }^2\mathrm{\theta }} \\ \overline{) \Rightarrow 1={\mathrm{e}}^2{\sec }^2\mathrm{\theta }} \\ \overline{) \Rightarrow {\cos }^2\mathrm{\theta }={\mathrm{e}}^2} \\ \overline{) \Rightarrow \mathrm{cos\theta }=\mathrm{e}} \end{gathered}$$