determine the correct value for each given description.\nthe least positive value k for which x = k is a…

Explanation:

Step1: Recall secant function definition

Recall that $\sec x=\frac{1}{\cos x}$. Vertical asymptotes occur where $\cos x = 0$. The general solutions of $\cos x=0$ are $x=\frac{\pi}{2}+n\pi,n\in\mathbb{Z}$. The least - positive value for a vertical asymptote is when $n = 0$, so $x=\frac{\pi}{2}$.

Step2: Recall cosecant function definition

Recall that $\csc x=\frac{1}{\sin x}$. Vertical asymptotes occur where $\sin x = 0$. The general solutions of $\sin x = 0$ are $x=n\pi,n\in\mathbb{Z}$. The least - positive value for a vertical asymptote is when $n = 1$, so $x=\pi$.

Step3: Analyze range of secant function

Since $\sec x=\frac{1}{\cos x}$ and $- 1\leqslant\cos x\leqslant1,\cos x\neq0$, the range of $\sec x$ is $(-\infty,-1]\cup[1,\infty)$. The least positive value in the range of $\sec x$ is $1$.

Step4: Analyze range of cosecant function

Since $\csc x=\frac{1}{\sin x}$ and $-1\leqslant\sin x\leqslant1,\sin x\neq0$, the range of $\csc x$ is $(-\infty,-1]\cup[1,\infty)$. The greatest negative value in the range of $\csc x$ is $-1$.

Step5: Solve $\sec x=-1$

Since $\sec x=\frac{1}{\cos x}=-1$, then $\cos x=-1$. The general solutions are $x=(2n + 1)\pi,n\in\mathbb{Z}$. The greatest negative value of $x$ for which $\sec x=-1$ is $x =-\pi$.

Step6: Solve $\csc x=-1$

Since $\csc x=\frac{1}{\sin x}=-1$, then $\sin x=-1$. The general solutions are $x=\frac{3\pi}{2}+2n\pi,n\in\mathbb{Z}$. The least positive value of $x$ for which $\csc x=-1$ is $x=\frac{3\pi}{2}$.

Answer:

The least positive value $k$ for which $x = k$ is a vertical asymptote for $y=\sec x$: $\frac{\pi}{2}$ The least positive value $k$ for which $x = k$ is a vertical asymptote for $y=\csc x$: $\pi$ The least positive value that is in the range of $y=\sec x$: $1$ The greatest negative value that is in the range of $y=\csc x$: $-1$ The greatest negative value of $x$ for which $\sec x=-1$: $-\pi$ The least positive value of $x$ for which $\csc x=-1$: $\frac{3\pi}{2}$