Answer
$$\cot t + t + \csc t + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dt}}{{1 - \sec t}}} \cr
& {\text{Multiply the numerator and the denomiantor by the conjugate of }}1 - \sec t \cr
& = \int {\frac{1}{{1 - \sec t}}} \left( {\frac{{1 + \sec t}}{{1 + \sec t}}} \right)dt \cr
& = \int {\frac{{1 + \sec t}}{{1 - {{\sec }^2}t}}} dt \cr
& {\text{Use the identity ta}}{{\text{n}}^2}t = {\sec ^2}t - 1 \cr
& = \int {\frac{{1 + \sec t}}{{ - {{\tan }^2}t}}} dt \cr
& {\text{distribute the numerator and simplify}} \cr
& = - \int {\left( {\frac{1}{{{{\tan }^2}t}} + \frac{{\sec t}}{{{{\tan }^2}t}}} \right)} dt \cr
& = - \int {\left( {{{\cot }^2}t + \left( {\frac{1}{{\cos t}}} \right)\left( {\frac{{{{\cos }^2}t}}{{{{\sin }^2}t}}} \right)} \right)} dt \cr
& {\text{Use the identity }}{\cot ^2}t = {\csc ^2}t - 1 \cr
& = - \int {\left( {{{\csc }^2}t - 1 + \left( {\frac{1}{{\sin t}}} \right)\left( {\frac{{\cos t}}{{\sin t}}} \right)} \right)} dt \cr
& = - \int {\left( {{{\csc }^2}t - 1 + \csc t\cot t} \right)} dt \cr
& = - \int {{{\csc }^2}t} dt + \int {dt} - \int {\csc t\cot t} dt \cr
& {\text{Integrating, we get}} \cr
& = - \left( { - \cot t} \right) + t - \left( { - \csc t} \right) + C \cr
& = \cot t + t + \csc t + C \cr} $$
