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OpenStudy (amyhashimoto77):
TRIG Convert the equation to polar form. 1. x^2-y^2 = 1
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OpenStudy (amyhashimoto77):
i got this so far. r^(2)cos^(2)theta - r^(2)sin^(2)theta = 1 r^(2)[cos^(2) theta - sin^(2)theta] = 1 r^(2) = 1/[cos^(2) theta - sin^(2)theta] r = √(1/[cos^(2) theta - sin^(2)theta]) r = 1/ √(cos^(2) theta - sin^(2)theta) Is this correct?
OpenStudy (amyhashimoto77):
@IrishBoy123 wait! I thought you were replying
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