a)

The polar coordinate \(\displaystyle{\left[\begin{array}{cc} {r}&\theta\end{array}\right]}\) corresponding to Cartesian coordinate \(\displaystyle{\left[\begin{array}{cc} {x}&\ {y}\end{array}\right]}{i}{s}{g}{i}{v}{e}{n}{b}{y}{\left[{r}=\sqrt{{{x}^{{{2}}}\ +\ {y}^{{{2}}}}}\ {\quad\text{and}\quad}\theta={{\tan}^{{-{1}}}{\left({\frac{{{y}}}{{{x}}}}\right)}}\right]}.\)

From the given data, \(\displaystyle{\left[{\left({x},\ {y}\right)}={\left(-{2},\ {2}\right)}\right]}.\)

Substitute \(\displaystyle{\left[{\left({x},\ {y}\right)}={\left(-{2},\ {2}\right)}\right]}\in{\left[{r}=\sqrt{{{x}^{{{2}}}\ +\ {y}^{{{2}}}}}\right]}\) and obtain the value of r.

\(\displaystyle{r}=\sqrt{{{2}^{{{2}}}\ +\ {\left(-{2}\right)}^{{{2}}}}}\)

\(\displaystyle=\pm\sqrt{{{8}}}\)

\(\displaystyle=\pm\ {2}\sqrt{{{2}}}\)

Similary, calculate the value of \(\displaystyle{\left[\theta\right]}\) as follows.

\(\displaystyle\theta={{\tan}^{{-{1}}}{\left(-{\frac{{{2}}}{{{2}}}}\right)}}\)

\(\displaystyle={{\tan}^{{-{1}}}{\left(-{1}\right)}}\)

\(\displaystyle={\frac{{{3}\pi}}{{{4}}}}\ +\pi{n},\ {n}\ \in{\mathbb{{Z}}}\)

For \(\displaystyle{\left[{r}={2}\sqrt{{{2}}}\ {\quad\text{and}\quad}\ {n}={0}\right]}{t}{h}{e}{p}{o}{l}{a}{r}{c}\infty{r}{d}\in{a}{t}{e}{i}{s}{\left[\begin{array}{cc} {2}\sqrt{{{2}}}&{\frac{{{3}\pi}}{{{4}}}}\end{array}\right]}.\)

For \(\displaystyle{\left[{r}={2}\sqrt{{{2}}}\ {\quad\text{and}\quad}\ {n}={2}\backslash\right]}{t}{h}{e}{p}{o}{l}{a}{r}{c}\infty{r}{d}\in{a}{t}{e}{i}{s}{\left[{\left({2}\sqrt{{{2}}},{\frac{{{11}\pi}}{{{4}}}}\right)}\backslash\right]}.\)

For \(\displaystyle{\left[{r}={2}\sqrt{{{2}}}\ {\quad\text{and}\quad}\ {n}=\ -{2}\backslash\right]}{t}{h}{e}{p}{o}{l}{a}{r}{c}\infty{r}{d}\in{a}{t}{e}{i}{s}{\left[{\left({2}\sqrt{{{2}}},\ -{\frac{{{5}\pi}}{{{4}}}}\right)}\backslash\right]}\).

For \(\displaystyle{\left[{r}={2}\sqrt{{{2}}}\ {\quad\text{and}\quad}\ {n}=\ -{1}\backslash\right]}{t}{h}{e}{p}{o}{l}{a}{r}{c}\infty{r}{d}\in{a}{t}{e}{i}{s}{\left[{\left(-{2}\sqrt{{{2}}},\ -{\frac{{\pi}}{{{4}}}}\right)}\backslash\right]}\)

Hence the given stament is true.

b) Note that the Cartesian coordinates corresponding to polar coordinates \(\displaystyle{\left[\begin{array}{cc} {r}&\theta\end{array}\right]}\) are given by \(\displaystyle{\left[{x}={r}{\cos{\theta}}\ {\quad\text{and}\quad}\ {y}={r}{\sin{\theta}}\right]}.\)

From the given data it can be concluded that,

\(\displaystyle{x}={r}{\cos{\theta}}\)

\(\displaystyle={4}\)

\(\displaystyle{y}={r}{\sin{\theta}}\)

\(\displaystyle=\ -{2}\)

Which is noting but a single point \(\displaystyle{\left[\begin{array}{cc} {4}&\ -{2}\end{array}\right]}\).

Therefore, the given statement is true.

c)

The given polar equations are \(\displaystyle{\left[{r}={2}\ {\quad\text{and}\quad}\ \theta={\frac{{\pi}}{{{4}}}}\right]}\).

Note that \(\displaystyle{\left[{r}={2}\right]}\) is an equation of a circle centered at the origin and having radius 2 while \(\displaystyle{\left[\theta={\frac{{\pi}}{{{4}}}}\right]}\) is an equation of a straight line passing through the origin.

Thus the line \(\displaystyle{\left[\theta={\frac{{\pi}}{{{4}}}}\right]}\) intersects the circle \(\displaystyle{r}={2}\) twise.

Therefore, the statement is false.

d)

The given point is \(\displaystyle{\left[{\left({3},{\frac{{\pi}}{{{2}}}}\right)}={\left({r},\theta\right)}\right]}\).

Change the radial coordinate to -3 and subtract \(\displaystyle{\left[\pi\right]}\) from the angle to obtain the alternate representation of the point as follows.

\(\displaystyle{\left[{\left(-{3},{\frac{{\pi}}{{{2}}}}\ -\pi\right)}={\left(-{3},-{\frac{{\pi}}{{{2}}}}\right)}\right]}\)

Thus, point \(\displaystyle{\left[\begin{array}{cc} -{3}&-{\frac{{\pi}}{{{2}}}}\end{array}\right]}{\quad\text{and}\quad}{\left[\begin{array}{cc} {3}&{\frac{{\pi}}{{{2}}}}\end{array}\right]}\) represent the same point.

Substitute \(\displaystyle{\left[{r}=\ -{3}\ {\quad\text{and}\quad}\ \theta=\ -{\frac{{\pi}}{{{2}}}}\ \in\ {r}={3}{\cos{{2}}}\theta\right]}\)

\(\displaystyle{\left[-{3}={3}{\cos{\ }}{2}{\left(-{\frac{{\pi}}{{{2}}}}\right)}\right.}\)

\(\displaystyle={3}{\cos{\ }}{\left(-\pi\right)}\)

\(\displaystyle={3}{\cos{\pi}}\)

\(\displaystyle=-{3}\)

Therefore, the point \(\displaystyle{\left[\begin{array}{cc} -{3}&-{\frac{{\pi}}{{{2}}}}\end{array}\right]}\) lies on the graph of \(\displaystyle{\left[{r}={3}{\cos{{2}}}\theta\right]}\) that is the point \(\displaystyle{\left[\begin{array}{cc} {3}&{\frac{{\pi}}{{{2}}}}\end{array}\right]}\) also lies on the graph of \(\displaystyle{\left[{r}={3}{\cos{{2}}}\theta\right]}\).

Therefore, the statement is true.

e)

The given equations are \(\displaystyle{\left[{r}={2}{\sec{\theta}}\right]}{\quad\text{and}\quad}{\left[{r}={3}{\csc{\theta}}\right]}.\)

Rewrite \(\displaystyle{\left[{r}={2}{\sec{\theta}}\right]}{\quad\text{and}\quad}{\left[{r}={3}{\csc{\theta}}\right]}\) as follows.

\(\displaystyle{r}={2}{\sec{\theta}}\)

\(\displaystyle{r}{\cos{\theta}}={2}\)

\(\displaystyle{r}={3}{\csc{\theta}}\)

\(\displaystyle{r}{\sin{\theta}}={3}\)

Note that \(\displaystyle{r}{\cos{\theta}}={x}{]}{\quad\text{and}\quad}{\left[{r}{\sin{\theta}}={y}\right]}\).

Thus, the given equations \(\displaystyle{\left[{x}={2}\right]}{\quad\text{and}\quad}{\left[{y}={3}\right]}\) represent the lines in the xy - plane.

Therefore, the given statement is true.

The polar coordinate \(\displaystyle{\left[\begin{array}{cc} {r}&\theta\end{array}\right]}\) corresponding to Cartesian coordinate \(\displaystyle{\left[\begin{array}{cc} {x}&\ {y}\end{array}\right]}{i}{s}{g}{i}{v}{e}{n}{b}{y}{\left[{r}=\sqrt{{{x}^{{{2}}}\ +\ {y}^{{{2}}}}}\ {\quad\text{and}\quad}\theta={{\tan}^{{-{1}}}{\left({\frac{{{y}}}{{{x}}}}\right)}}\right]}.\)

From the given data, \(\displaystyle{\left[{\left({x},\ {y}\right)}={\left(-{2},\ {2}\right)}\right]}.\)

Substitute \(\displaystyle{\left[{\left({x},\ {y}\right)}={\left(-{2},\ {2}\right)}\right]}\in{\left[{r}=\sqrt{{{x}^{{{2}}}\ +\ {y}^{{{2}}}}}\right]}\) and obtain the value of r.

\(\displaystyle{r}=\sqrt{{{2}^{{{2}}}\ +\ {\left(-{2}\right)}^{{{2}}}}}\)

\(\displaystyle=\pm\sqrt{{{8}}}\)

\(\displaystyle=\pm\ {2}\sqrt{{{2}}}\)

Similary, calculate the value of \(\displaystyle{\left[\theta\right]}\) as follows.

\(\displaystyle\theta={{\tan}^{{-{1}}}{\left(-{\frac{{{2}}}{{{2}}}}\right)}}\)

\(\displaystyle={{\tan}^{{-{1}}}{\left(-{1}\right)}}\)

\(\displaystyle={\frac{{{3}\pi}}{{{4}}}}\ +\pi{n},\ {n}\ \in{\mathbb{{Z}}}\)

For \(\displaystyle{\left[{r}={2}\sqrt{{{2}}}\ {\quad\text{and}\quad}\ {n}={0}\right]}{t}{h}{e}{p}{o}{l}{a}{r}{c}\infty{r}{d}\in{a}{t}{e}{i}{s}{\left[\begin{array}{cc} {2}\sqrt{{{2}}}&{\frac{{{3}\pi}}{{{4}}}}\end{array}\right]}.\)

For \(\displaystyle{\left[{r}={2}\sqrt{{{2}}}\ {\quad\text{and}\quad}\ {n}={2}\backslash\right]}{t}{h}{e}{p}{o}{l}{a}{r}{c}\infty{r}{d}\in{a}{t}{e}{i}{s}{\left[{\left({2}\sqrt{{{2}}},{\frac{{{11}\pi}}{{{4}}}}\right)}\backslash\right]}.\)

For \(\displaystyle{\left[{r}={2}\sqrt{{{2}}}\ {\quad\text{and}\quad}\ {n}=\ -{2}\backslash\right]}{t}{h}{e}{p}{o}{l}{a}{r}{c}\infty{r}{d}\in{a}{t}{e}{i}{s}{\left[{\left({2}\sqrt{{{2}}},\ -{\frac{{{5}\pi}}{{{4}}}}\right)}\backslash\right]}\).

For \(\displaystyle{\left[{r}={2}\sqrt{{{2}}}\ {\quad\text{and}\quad}\ {n}=\ -{1}\backslash\right]}{t}{h}{e}{p}{o}{l}{a}{r}{c}\infty{r}{d}\in{a}{t}{e}{i}{s}{\left[{\left(-{2}\sqrt{{{2}}},\ -{\frac{{\pi}}{{{4}}}}\right)}\backslash\right]}\)

Hence the given stament is true.

b) Note that the Cartesian coordinates corresponding to polar coordinates \(\displaystyle{\left[\begin{array}{cc} {r}&\theta\end{array}\right]}\) are given by \(\displaystyle{\left[{x}={r}{\cos{\theta}}\ {\quad\text{and}\quad}\ {y}={r}{\sin{\theta}}\right]}.\)

From the given data it can be concluded that,

\(\displaystyle{x}={r}{\cos{\theta}}\)

\(\displaystyle={4}\)

\(\displaystyle{y}={r}{\sin{\theta}}\)

\(\displaystyle=\ -{2}\)

Which is noting but a single point \(\displaystyle{\left[\begin{array}{cc} {4}&\ -{2}\end{array}\right]}\).

Therefore, the given statement is true.

c)

The given polar equations are \(\displaystyle{\left[{r}={2}\ {\quad\text{and}\quad}\ \theta={\frac{{\pi}}{{{4}}}}\right]}\).

Note that \(\displaystyle{\left[{r}={2}\right]}\) is an equation of a circle centered at the origin and having radius 2 while \(\displaystyle{\left[\theta={\frac{{\pi}}{{{4}}}}\right]}\) is an equation of a straight line passing through the origin.

Thus the line \(\displaystyle{\left[\theta={\frac{{\pi}}{{{4}}}}\right]}\) intersects the circle \(\displaystyle{r}={2}\) twise.

Therefore, the statement is false.

d)

The given point is \(\displaystyle{\left[{\left({3},{\frac{{\pi}}{{{2}}}}\right)}={\left({r},\theta\right)}\right]}\).

Change the radial coordinate to -3 and subtract \(\displaystyle{\left[\pi\right]}\) from the angle to obtain the alternate representation of the point as follows.

\(\displaystyle{\left[{\left(-{3},{\frac{{\pi}}{{{2}}}}\ -\pi\right)}={\left(-{3},-{\frac{{\pi}}{{{2}}}}\right)}\right]}\)

Thus, point \(\displaystyle{\left[\begin{array}{cc} -{3}&-{\frac{{\pi}}{{{2}}}}\end{array}\right]}{\quad\text{and}\quad}{\left[\begin{array}{cc} {3}&{\frac{{\pi}}{{{2}}}}\end{array}\right]}\) represent the same point.

Substitute \(\displaystyle{\left[{r}=\ -{3}\ {\quad\text{and}\quad}\ \theta=\ -{\frac{{\pi}}{{{2}}}}\ \in\ {r}={3}{\cos{{2}}}\theta\right]}\)

\(\displaystyle{\left[-{3}={3}{\cos{\ }}{2}{\left(-{\frac{{\pi}}{{{2}}}}\right)}\right.}\)

\(\displaystyle={3}{\cos{\ }}{\left(-\pi\right)}\)

\(\displaystyle={3}{\cos{\pi}}\)

\(\displaystyle=-{3}\)

Therefore, the point \(\displaystyle{\left[\begin{array}{cc} -{3}&-{\frac{{\pi}}{{{2}}}}\end{array}\right]}\) lies on the graph of \(\displaystyle{\left[{r}={3}{\cos{{2}}}\theta\right]}\) that is the point \(\displaystyle{\left[\begin{array}{cc} {3}&{\frac{{\pi}}{{{2}}}}\end{array}\right]}\) also lies on the graph of \(\displaystyle{\left[{r}={3}{\cos{{2}}}\theta\right]}\).

Therefore, the statement is true.

e)

The given equations are \(\displaystyle{\left[{r}={2}{\sec{\theta}}\right]}{\quad\text{and}\quad}{\left[{r}={3}{\csc{\theta}}\right]}.\)

Rewrite \(\displaystyle{\left[{r}={2}{\sec{\theta}}\right]}{\quad\text{and}\quad}{\left[{r}={3}{\csc{\theta}}\right]}\) as follows.

\(\displaystyle{r}={2}{\sec{\theta}}\)

\(\displaystyle{r}{\cos{\theta}}={2}\)

\(\displaystyle{r}={3}{\csc{\theta}}\)

\(\displaystyle{r}{\sin{\theta}}={3}\)

Note that \(\displaystyle{r}{\cos{\theta}}={x}{]}{\quad\text{and}\quad}{\left[{r}{\sin{\theta}}={y}\right]}\).

Thus, the given equations \(\displaystyle{\left[{x}={2}\right]}{\quad\text{and}\quad}{\left[{y}={3}\right]}\) represent the lines in the xy - plane.

Therefore, the given statement is true.