# Polar to rectangular form

• Dec 15th 2008, 07:44 PM
cmr7
Polar to rectangular form
r^2= sec^2(theta) - tan^2(theta)
• Dec 15th 2008, 07:48 PM
Mathstud28
$\displaystyle \sec(x)\ne\frac{1}{\sin(x)}$ :(
• Dec 15th 2008, 07:56 PM
cmr7
Thank you!!!!!
• Dec 15th 2008, 08:02 PM
Soroban
Hello, cmr7!

Someone created this problem as a joke . . .

Quote:

Write in rectangular form: .$\displaystyle r^2\:=\: \sec^2\!\theta - \tan^2\!\theta$

We're expected to know this identity: .$\displaystyle \sec^2\!A - \tan^2\!A \:=\:1$

. . and this conversion: .$\displaystyle r^2 \:=\:x^2+y^2$

So: .$\displaystyle r^2 \:=\:\sec^2\!\theta - \tan^2\!\theta\;\text{ becomes: }\:r^2 \:=\:1 \quad\Rightarrow\quad x^2+y^2 \:=\:1$

• Dec 15th 2008, 08:06 PM
Mathstud28
Quote:

Originally Posted by Soroban
Hello, cmr7!

Someone created this problem as a joke . . .

Yeah, apparently for me.
• Dec 15th 2008, 08:08 PM
cmr7
me too
I am in the class right now so all this should be fresh on my mind and it got past me also. I am going on 4 night sin a row of studying for this final. Maybe I should just get some sleep now haha!