The value of ${\int }_{1/e}^{\tan x} \frac{t}{1+t^2}\mathrm{d}t+{\int }_{1/e}^{\cot x} \frac{\mathrm{d}t}{t\left(1+t^2\right)}$ is

$\frac{1}{t\left(1+t^2\right)}=\frac{1}{t}-\frac{t}{1+t^2}$So the required integral is  equal to

$\begin{array}{l}{\int }_{\frac{1}{e}}^{\tan x} \frac{tdx}{1+t^2}+{\int }_{\frac{1}{e}}^{\cot x} \frac{dt}{t}-{\int }_{\frac{1}{e}}^{\cot x} \frac{tdt}{1+t^2}\\ ={\left.\frac{1}{2}\log \left(1+t^2\right)\right|}_{\frac{1}{e}}^{\tan x}+{\left.\log t\right|}_{\frac{1}{e}}^{\cot x}-{\left.\frac{1}{2}\log \left(1+t^2\right)\right|}_{\frac{1}{e}}^{\cot x}\\ =\log |\sec x|+\log |\cot x|+1-\log |\mathrm{cosec}x|=1\end{array}$