A pair of straight lines is passing through the point (1, 1). One of the lines makes an angle $\theta with$ the positive direction of X- axis and the other makes the same angle with the positive direction of Y-axis. If the equation of the pair of straight lines is $x^2-\left(a+2\right)xy+y^2+a\left(x+y-1\right)=0,a\ne -2thenthevalueof\theta is$

The lines will be

$\begin{array}{l}y-1=\mathrm{Tan}\theta \left(x-1\right)\\ \Rightarrow \tan \theta \left(x-1\right)-\left(y-1\right)=0\\ andy-1=\cot \theta \left(x-1\right)\\ \Rightarrow \cot \theta \left(x-1\right)-\left(y-1\right)=0\\ combainedequationis\left(\mathrm{Tan}\theta \left(x-1\right)-\left(y-1\right)\right)\left(\cot \theta \left(x-1\right)-\left(y-1\right)\right)=0\\ \Rightarrow {\left(x-1\right)}^2-\mathrm{Tan}\theta \left(x-1\right)\left(y-1\right)-\cot \theta \left(x-1\right)\left(y-1\right)+{\left(y-1\right)}^2=0\\ \Rightarrow x^2+1-2x-\left(\tan \theta +\cot \theta \right)\left(x-1\right)\left(y-1\right)+y^2+1-2y=0\\ \Rightarrow x^2-\left(\tan \theta +\cot \theta \right)xy+y^2-2x-2y+3+\left(\tan \theta +\cot \theta \right)\left(x+y\right)=0\\ Comparingwith{x}^2-\left(a+2\right)xy+y^2+a\left(x+y-1\right)=0\\ \therefore \mathrm{Tan}\theta +\cot \theta =a+2\\ \Rightarrow \frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }=a+2\end{array}$

$\begin{array}{l}\Rightarrow \frac{1}{\sin \theta \cos \theta }=a+2\\ \Rightarrow \frac{1}{2\sin \theta \cos \theta }=\frac{a+2}{2}\\ \Rightarrow \frac{1}{\sin 2\theta }=\frac{a+2}{2}\\ \Rightarrow \sin 2\theta =\frac{2}{a+2}\\ \Rightarrow 2\theta ={\sin }^{-1}\left(\frac{2}{a+2}\right)\\ \Rightarrow \theta =\frac{1}{2}{\sin }^{-1}\left(\frac{2}{a+2}\right)\end{array}$