Question

Find the sign of the expression if the terminal point determined by $t$is in the given quadrant.

$\tan t\sec t,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Quadrant\hspace{0.17em}\hspace{0.17em}IV}$

Short answer

In quadrant IV, $\mathbf{tan}\mathit{t}\mathbf{sec}\mathit{t}\mathbf{<}\mathbf{0}$.

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 obtained in this way is called the terminal point determined by the real number $t$.

Step 2. Find the sign-in quadrant IV. 

Determine the sign of the trigonometric function of $t$by noting the quadrant in which the terminal point lies. The terminal point of $t$lies in quadrant IV and in quadrant IV $x$-coordinate is positive and$y$-coordinate is negative.

Step 3. Find the sign of the expression.

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 $t$.

Since $x>0$, therefore, $\cos t>0$, which implies

$$\begin{gathered} \sec t=\frac{1}{\cos t} \\ >0 \end{gathered}$$

and because $y<0$, therefore, $\sin t<0$.

$$\begin{gathered} \tan t=\frac{\sin t}{\cos t} \\ <0 \end{gathered}$$

$$\begin{gathered} \tan t\sec t=\frac{\sin t}{\cos t}\times \frac{1}{\cos t} \\ <0 \end{gathered}$$

Hence, in quadrant IV, $\tan t\sec t<0$.