Cartesian coordinate of point is (1 + √3, 1 - √3) = (x, y) (Let)
\(\therefore\) x = 1 + √3 & y = 1 - √3
\(\because\) r = \(\sqrt{x^2+y^2}=\sqrt{(1+\sqrt3)^2+(1-\sqrt3)^2}\)
= \(\sqrt{(1 + 3 + 2\sqrt3)+(1 + 3 - 2\sqrt3)}\)
= \(\sqrt{4+4}\) = 2√2
and θ = tan-1 y/x = tan-1\(\left(\frac{1-\sqrt3}{1+\sqrt3}\right)\)
= tan-1\(\left(\frac{tan45^o-tan60^o}{1+tan45^o\,tan60^o}\right)\)
\((\because tan45^o = 1 \,\&\,tan60^o = \sqrt3)\)
= tan-1(tan(45° - 60°))
\(\left(\because tan(A-B)=\frac{tanA-tanB}{1+tan AtanB}\right)\)
= tan-1(tan(-15°))
= tan-1(-tan 15°)
(\(\because\) tan(-θ) = -tan θ)
= tan-1(tan(180°- 15°))
(\(\because\) tan(180°- θ) = -tan θ)
= tan-1(tan 165°)
= 165°
Hence, polar coordinates of given point are (r, θ) = (2√2, 165°)
