Question

From the information given, find the quadrant in which the terminal point is determined by $t$lies.

$\csc t>0\text{and}\sec t<0$

Short answer

The quadrant in which the terminal point is determined by $t$lies isquadrant II.

Step-by-step solution

Step 1. State the definition of the terminal point on the unit circle.

Starting at the point $(1,0)$, if $t\ge 0$, then $t$is the distance along the unit circle in a clockwise direction and if $t<0$, then $\left|t\right|$is the distance along the unit circle in a clockwise direction. Here, $t$is a real number and this distance generates a point $P\left(x,y\right)$ on the unit circle. The point $P\left(x,y\right)$ obtained in this way is called the terminal point determined by the real number $t$.

Step 2. Find the sign of x and y coordinates. 

Determine the signs of $x$and $y$coordinates by observing the signs of trigonometric functions.

Note that for the given point $t$on the unit circle $\cos t=x$, $\sin t=y$and $\tan t=\frac{y}{x}$where $\left(x,y\right)$is the terminal point to a real number $t$.

$$\begin{gathered} \csc t=\frac{1}{\sin t} \\ =\frac{1}{y} \\ y>0\left(\because \csc t>0\right) \end{gathered}$$

And,

$$\begin{gathered} \sec t=\frac{1}{\cos t} \\ =\frac{1}{x} \\ x<0\left(\because \sec t<0\right) \end{gathered}$$

Therefore, $x$coordinate is negative, and $y$coordinate is positive.

Step 3. Find the quadrant.

Since $x$ coordinate is negative and $y$ coordinate is positive, it implies that the terminal point lies in quadrant II.