If $x\cos \theta +y\sin \theta =p$is the normal form and y= mx +c is the slope - intercept form of the line x +2y +1 =0, then the value of ${\tan }^{-1}(\tan \theta +m+c)$ in degrees is ……

$x+2y+1=0\Rightarrow -2y=x+1\Rightarrow y=\left(\frac{{\displaystyle -1}}{{\displaystyle 2}}\right)x+\left(\frac{{\displaystyle -1}}{{\displaystyle 2}}\right)$ which is the form y = mx +c

$\therefore m=-1/2,c=-1/2$

$x+2y+1=0\Rightarrow -x-2y=1\Rightarrow \left(\frac{{\displaystyle -1}}{{\displaystyle \sqrt{5}}}\right)x+\left(\frac{{\displaystyle -2}}{{\displaystyle \sqrt{5}}}\right)=\frac{1}{\sqrt{5}}$

$\Rightarrow x\cos \theta +y\sin \theta =p$where $\cos \theta =\frac{-1}{\sqrt{5}},\sin \theta =\frac{-2}{\sqrt{5}}\Rightarrow \tan \theta =2$

$Ta{n}^{-1}(\tan \theta +m+c)=Ta{n}^{-1}\left(2-\frac{{\displaystyle 1}}{{\displaystyle 2}}-\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)=Ta{n}^{-1}\left(1\right)=\frac{\pi }{4}=45°$