Chapter 1 - Trigonometric Functions - Section 1.4 Using the Definitions of the Trigonometric Functions - 1.4 Exercises - Page 38: 71

1.05

Recall the identity $1+\cot^{2}\theta=\csc^{2}\theta$ Knowing that $\csc\theta=-1.45$, we get $1+\cot^{2}\theta=(-1.45)^{2}=2.1025$ $\implies \cot^{2}\theta=2.1025-1=1.1025$ Or $\cot\theta=\pm \sqrt {1.1025}=\pm 1.05$ In the third quadrant, $\cot\theta$ is positive. Therefore, $\cot\theta=1.05$