Question

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

$\frac{\tan t\sin t}{\cot t},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Quadrant\hspace{0.17em}\hspace{0.17em}III}$

Short answer

In quadrant III,$\frac{\mathbf{tan}\mathbf{t}\mathbf{sin}\mathbf{t}}{\mathbf{cot}\mathbf{t}}\mathbf{<}\mathbf{0}$

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 $P\left(x,y\right)$obtained in this way is called the terminal point determined by the real number $t$.

Step 2. Find the sign-in quadrant II. 

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 III and in quadrant III $x$-coordinate is negative and$y$-coordinate is also 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, $\cos t<0$and because $y<0$, therefore, $\sin t<0$.

$$\begin{gathered} \frac{\tan t\sin t}{\cot t}=\frac{\left(\frac{\sin t}{\cos t}\right)\sin t}{\left(\frac{\cos t}{\sin t}\right)} \\ =\frac{{\sin }^{3}t}{{\cos }^{2}t} \\ <0\left({\cos }^{2}t>0,{\sin }^{3}t<0\right) \end{gathered}$$

Hence, in quadrant III, $\frac{\tan t\sin t}{\cot t}<0$.