The general solution of $\mathrm{sec\theta }=$$\frac{{\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(2\right)+{\tan }^{-1}\left(3\right)}{{\tan }^{-1}\left(1\right)+{\tan }^{-1}(1/2)+{\tan }^{-1}(1/3)}\mathrm{is} \mathrm{\theta }=$

 

$$\begin{gathered} {\tan }^{-1}2+{\tan }^{-1}3=\mathrm{\pi }+{\tan }^{-1}\left(\frac{2+3}{1-6}\right)=\mathrm{\pi }-\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.=\raisebox{1ex}{$3\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right. \\ {\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3} \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{1}{2}+\frac{1}{3}}}{1-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}\right) ={\tan }^{-1}\left(1\right) =\frac{\mathrm{\pi }}{4} \\ \frac{{\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(2\right)+{\tan }^{-1}\left(3\right)}{{\tan }^{-1}\left(1\right)+{\tan }^{-1}(1/2)+{\tan }^{-1}(1/3)}=\frac{{\displaystyle \frac{\mathrm{\pi }}{4}}+{\displaystyle \frac{3\mathrm{\pi }}{4}}}{{\displaystyle \frac{\mathrm{\pi }}{4}}+{\displaystyle \frac{\mathrm{\pi }}{4}}} =\frac{\mathrm{\pi }}{{\displaystyle \frac{\mathrm{\pi }}{2}}}=2 \\ \Rightarrow \sec \mathrm{\theta }=2 \Rightarrow \cos \mathrm{\theta }=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ \mathrm{\theta }=2\mathrm{n\pi }\pm \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \end{gathered}$$