Answer
$$\frac{{dy}}{{d\theta }} = - 1$$
Work Step by Step
$$\eqalign{
& y = {\csc ^{ - 1}}\left( {\sec \theta } \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 < \theta < \pi /2 \cr
& {\text{find the derivative of }}y{\text{ with respect to }}\theta \cr
& \frac{{dy}}{{d\theta }} = \frac{{d\left( {{{\csc }^{ - 1}}\left( {\sec \theta } \right)} \right)}}{{d\theta }} \cr
& {\text{use the formula }}\cr
&\frac{{d\left( {{{\csc }^{ - 1}}u} \right)}}{{d\theta }} = - \frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }}\frac{{du}}{{d\theta }}{\text{ }}\left| u \right|{\text{ > 1 }}\cr
&\left( {{\text{see page 418}}} \right) \cr
& \frac{{dy}}{{d\theta }} = - \frac{1}{{\left| {\sec \theta } \right|\sqrt {{{\left( {\sec \theta } \right)}^2} - 1} }}\frac{{d\left( {\sec \theta } \right)}}{{d\theta }} \cr
& \frac{{dy}}{{d\theta }} = - \frac{1}{{\left| {\sec \theta } \right|\sqrt {{{\sec }^2}\theta - 1} }}\left( {\sec \theta \tan \theta } \right) \cr
& {\text{simplifying, we get:}} \cr
& \frac{{dy}}{{d\theta }} = - \frac{{\sec \theta \tan \theta }}{{\left| {\sec \theta } \right|\sqrt {{{\tan }^2}\theta } }} \cr
& \frac{{dy}}{{d\theta }} = - \frac{{\sec \theta \tan \theta }}{{\left| {\sec \theta } \right|\tan \theta }} \cr
& \frac{{dy}}{{d\theta }} = - \frac{{\sec \theta }}{{\left| {\sec \theta } \right|}} \cr
& \sec \theta {\text{ is positive for the interval }}\cr
&0 \lt \theta \lt \pi /2{\text{, thus:}} \cr
& \frac{{dy}}{{d\theta }} = - \frac{{\sec \theta }}{{\sec \theta }} \cr
& \frac{{dy}}{{d\theta }} = - 1 \cr} $$
