Question

Write the first expression in terms of the second if the terminal point determined by $t$is in the given quadrant.

$\csc t,\cot t;\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Quadrant\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}III}$

Short answer

The first expression in terms of the second if the terminal point determined by $t$is in the given quadrant is$\mathit{c}\mathit{s}\mathit{c}\mathit{t}\mathbf{=}\mathbf{-}\sqrt{\mathbf{1}\mathbf{+}\mathbf{c}\mathbf{o}{\mathbf{t}}^{\mathbf{2}}\mathbf{t}}$.

Step-by-step solution

Step 1. State the Trigonometric identities.

The reciprocal identities are $\csc t=\frac{1}{\sin t}$, $\sec t=\frac{1}{\cos t}$, $\cot t=\frac{1}{\tan t}$, $\tan t=\frac{\sin t}{\cos t}$ and $\cot =\frac{\cos t}{\sin t}$.

The Pythagorean identities are:

1. ${\sin }^{2}t+{\cos }^{2}t=1$

2. ${\tan }^{2}t+1={\sec }^{2}t$

3.${\cot }^{2}t+1={\csc }^{2}t$

Step 2. Formula used. 

Now use Pythagorean Identity ${\cot }^{2}t+1={\csc }^{2}t$, to rewrite the expression $\sec t$in the terms of $\tan t$.

$$\begin{gathered} {\cot }^{2}t+1={\csc }^{2}t \\ \csc t=\pm \sqrt{1+{\cot }^{2}t}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Takesquarerootbothsides} \end{gathered}$$

Step 3. State the conclusion.

Now, $t$is given to be in quadrant III, and quadrant III, $\cot >0$and $\csc t<0$ therefore,$\csc t=\sqrt{1+{\cot }^{2}t}$ is not a valid expression as the overall expression must be negative but here it will be positive. So, $\csc t=-\sqrt{1+{\cot }^{2}t}$.