What is the Cartesian form of ${(r + 3)}^{2} = {tan}^{2} θ - {sec}^{2} θ$ $(r+3)^2 = tan^2theta-sec^2theta$ ?

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Answer 1 · 2017-01-25T06:15:17

There is no real graph for this equation.

$0 \leq {(r + 3)}^{2} = - 1$ $0<=(r+3)^2= -1$ , using s $e c^{2} θ - {tan}^{2} θ = 1$ $ec^2theta-tan^2theta=1$

Indeed, an invalid inequality.

Such examples for conversions could be avoided.

What is the Cartesian form of (r+3)^2 = tan^2theta-sec^2theta (r+3)2=tan2θ−sec2θ(r+3)^2 = tan^2theta-sec^2theta ?