Chapter 6, Problem 1RE

To determine

To find:

Remaining trigonometric ratio based on the information given.

Explanation of Solution

Given:

$\sin \theta =-\frac{2}{3}\text{ and }\cos \theta <0$

Concept used: signs of value of the six trigonometric ratios in various quadrants can be represented, by the following diagram.

Calculation:

We Observe that the value of $\sin \theta$ is negative and $\cos \theta$ is also negative this means $\theta$ lies in the 3^rd quadrant.

$\begin{array}{l}\cos \theta =\sqrt{1-{\left(\sin \theta \right)}^2}=-\sqrt{1-{\left(-\frac{2}{3}\right)}^2}=\frac{-\sqrt{5}}{3}\\ \tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{-\frac{2}{3}}{\frac{-\sqrt{5}}{3}}=\frac{2\sqrt{5}}{5}\\ \Rightarrow \cot \theta =\frac{\cos \theta }{\sin \theta }=\frac{\sqrt{5}}{2}\\ \Rightarrow \sec \theta =\frac{1}{\cos \theta }=\frac{-3\sqrt{5}}{5}\\ \Rightarrow \text{csc}\theta =\frac{1}{\sin \theta }=\frac{-3}{2}\end{array}$.