If the pair of lines given by  $\left(x^2+y^2\right){\cos }^2\theta ={\left(x\cos \theta +y\sin \theta \right)}^2$are perpendicular to each other, then $\theta =$ 

Given Equation is $\left(x^2+y^2\right){\cos }^2\theta ={\left(x\cos \theta +y\sin \theta \right)}^2$

$\begin{array}{l}\Rightarrow x^2{\cos }^2\theta +y^2{\cos }^2\theta =x^2{\cos }^2\theta +y^2{\sin }^2\theta +2xy\cos \theta \sin \theta \\ \Rightarrow 2xy\cos \theta \sin \theta +\left({\sin }^2\theta -{\cos }^2\theta \right)y^2=0\end{array}$

Comparing with $a{x}^2+2hxy+b{y}^2=0$

$a=0,h=\cos \theta \sin \theta ,b={\sin }^2\theta -{\cos }^2\theta$

The lines are $\perp r$ iff a+b=0

$\begin{array}{l}\Rightarrow {\sin }^2\theta -{\cos }^2\theta =0\\ \Rightarrow {\mathrm{Tan}}^2\theta =1\\ \Rightarrow \mathrm{Tan}\theta =1\\ \Rightarrow \theta =\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.\end{array}$