Start by multiplying both sides by \(\displaystyle r\).

\(\displaystyle r=4\cos\theta\)

\(\displaystyle r^2=4r\cos\theta\)

Keep in mind that \(\displaystyle r^2=x^2+y^2\)

\(\displaystyle x^2+y^2=4r\cos\theta\)

Remember that \(\displaystyle x=r\cos\theta\)

So then,

\(\displaystyle x^2+y^2=4x\)

\(\displaystyle x^2-4x+y^2=0\)

Now, complete the square.

\(\displaystyle (x^2-4x+4)+y^2=4\)

\(\displaystyle (x-2)^2+y^2=4\)

Start by multiplying both sides by \(\displaystyle r\).

\(\displaystyle r=8\sin\theta\)

\(\displaystyle r^2=8r\sin\theta\)

Remember that \(\displaystyle r^2=x^2+y^2\)

\(\displaystyle x^2+y^2=8r\sin\theta\)

Keep in mind that \(\displaystyle y=r\sin\theta\)

So then,

\(\displaystyle x^2+y^2=8y\)

\(\displaystyle x^2+y^2-8x=0\)

Now, complete the square.

\(\displaystyle x^2+(y^2-8x+16)=16\)

\(\displaystyle x^2+(y-4)^2=16\)

This is a graph of a circle with a radius of \(\displaystyle 4\) and a center at \(\displaystyle (0, 4)\)

Remember that \(\displaystyle \sec\theta=\frac{1}{\cos\theta}\)

So then \(\displaystyle r=8\sec\theta\) becomes

\(\displaystyle r=\frac{8}{\cos\theta}\)

Now, multiply both sides by \(\displaystyle \cos\theta\) to get rid of the fraction.

\(\displaystyle r\cos\theta=8\)

Since \(\displaystyle x=r\cos\theta,\) the rectangular form of this equation is

\(\displaystyle x=8\)

Start by using the double angle formula for \(\displaystyle \sin\).

\(\displaystyle \sin(2\theta)=2\sin\theta\cos\theta\)

Substitute that into the equation gives the following:

\(\displaystyle r^2=4(2\sin\theta\cos\theta)\)

\(\displaystyle r^2=8\sin\theta\cos\theta\)

Because we need \(\displaystyle r\cos\theta\) and \(\displaystyle r\sin\theta\) to get \(\displaystyle x\) and \(\displaystyle y\) respectively, multiply both sides by \(\displaystyle r^2\).

\(\displaystyle (r^2)(r^2)=8(r\sin\theta)(r\cos\theta)\)

Now, recall that \(\displaystyle r^2=x^2+y^2\).

\(\displaystyle (x^2+y^2)(x^2+y^2)=8xy\)

\(\displaystyle (x^2+y^2)^2=8xy\)

Start by taking the tangent.

\(\displaystyle \tan\theta=\tan\frac{5\pi}{6}\)

\(\displaystyle \tan\theta=-\frac{\sqrt3}{3}\)

Recall that \(\displaystyle \tan\theta=\frac{y}{x}\)

\(\displaystyle \frac{y}{x}=-\frac{\sqrt3}{3}\)

\(\displaystyle y=-\frac{\sqrt3}{3}x\)

Start by taking the tangent.

\(\displaystyle \tan\theta=\tan\frac{5\pi}{4}\)

\(\displaystyle \tan\theta=1\)

Recall that \(\displaystyle \tan\theta=\frac{y}{x}\)

\(\displaystyle \frac{y}{x}=1\)

\(\displaystyle y=x\)

Recall that \(\displaystyle \sec\theta=\frac{1}{\cos\theta}\)

Substituting this into the given equation gives

\(\displaystyle r=4\frac{1}{\cos\theta}\tan\theta\)

Multiply both sides by \(\displaystyle \cos\theta\) to get rid of the fraction.

\(\displaystyle r\cos\theta=4\tan\theta\)

Recall that \(\displaystyle x=r\cos\theta\) and that \(\displaystyle \tan\theta=\frac{y}{x}\)

\(\displaystyle x=4\frac{y}{x}\)

\(\displaystyle 4y=x^2\)

\(\displaystyle y=\frac{1}{4}x^2\)

Recall that \(\displaystyle \csc\theta=\frac{1}{sin\theta}\) and \(\displaystyle \cot\theta=\frac{1}{tan\theta}\)

\(\displaystyle r=2\frac{1}{\sin\theta}\frac{1}{\tan\theta}\)

Multiply both sides by \(\displaystyle \sin\theta\)

\(\displaystyle r\sin\theta=2(\frac{1}{\tan\theta})\)

Recall that \(\displaystyle \tan\theta=\frac{y}{x}\) and \(\displaystyle r\sin\theta=y\)

\(\displaystyle y=2(\frac{1}{\frac{y}{x}})\)

\(\displaystyle y=2(\frac{x}{y})\)

\(\displaystyle 2x=y^2\)

\(\displaystyle y=\sqrt{2x}\)

Recall that \(\displaystyle \sec\theta=\frac{1}{\cos\theta}\)

Plugging this into the equation gives us

\(\displaystyle r=-3\frac{1}{\cos\theta}\)

Multiply both sides by \(\displaystyle \cos\theta\) to get rid of the fraction.

\(\displaystyle r\cos\theta=-3\)

Recall that \(\displaystyle r\cos\theta=x\)

So then the rectangular form of the equation is \(\displaystyle x=-3\)

Recall that \(\displaystyle \csc\theta=\frac{1}{\sin\theta}\)

\(\displaystyle 6r=\frac{1}{\sin\theta}\)

Multiply both sides by \(\displaystyle \sin\theta\) to get rid of the fraction.

\(\displaystyle 6r\sin\theta=1\)

Recall that \(\displaystyle y=r\sin\theta\).

\(\displaystyle 6y=1\)

\(\displaystyle y=\frac{1}{6}\)