If the cartesian coordinates of a point are $(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}),$ then the polar coordinates are

The correct option is B $(1,-\frac{\pi }{4})$
Let $(x,y)\equiv (\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$
Since, $r=\sqrt{x^2+y^2}$
$\Rightarrow r=\sqrt{{\left(\frac{1}{\sqrt{2}}\right)}^2+{(-\frac{1}{\sqrt{2}})}^2}\Rightarrow r=\sqrt{\frac{1}{2}+\frac{1}{2}}=1$

Also, $(x,y)\equiv (\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$ lies in $4^{\text{th}}$ quadrant.
$\therefore \theta =-{\tan }^{-1}\left|\frac{y}{x}\right|\Rightarrow \theta =-{\tan }^{-1}|-\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{1}|\Rightarrow \theta =-{\tan }^{-1}\left(1\right)\Rightarrow \theta =-\frac{\pi }{4}\therefore (r,\theta )\equiv (1,-\frac{\pi }{4})$