If xcos$\mathrm{\theta }$+ysin$\mathrm{\theta }$=p is the normal form and y=mx +c is the slope - intercept form of the line x+2y+ 1=0, then Tan^-1(tan$\mathrm{\theta }$ + m+ c)=

$\mathrm{x}+2\mathrm{y}+1=0\Rightarrow -2\mathrm{y}=\mathrm{x}+1\Rightarrow \mathrm{y}=\left(\frac{-1}{2}\right)\mathrm{x}+\left(\frac{-1}{2}\right)$which is  in  the form y=mx+ c
$\therefore \mathrm{m}=-\frac{1}{2},\mathrm{c}=-\frac{1}{2}$
$\mathrm{x}+2\mathrm{y}+1=0\Rightarrow -\mathrm{x}-2\mathrm{y}=1\Rightarrow \left(\frac{-1}{\sqrt{5}}\right)\mathrm{x}+\left(\frac{-2}{\sqrt{5}}\right)y=\frac{1}{\sqrt{5}}$
$\begin{array}{l}\Rightarrow \mathrm{xcos}\mathrm{\theta }+\mathrm{ysin}\mathrm{\theta }=\mathrm{p} \mathrm{where} \cos \mathrm{\theta }=\frac{-1}{\sqrt{5}},\sin \mathrm{\theta }=\frac{-2}{\sqrt{5}}\Rightarrow \tan \mathrm{\theta }=2\\ {\mathrm{Tan}}^{-1}(\tan \mathrm{\theta }+\mathrm{m}+\mathrm{c})={\mathrm{Tan}}^{-1}\left(2-\frac{1}{2}-\frac{1}{2}\right)={\mathrm{Tan}}^{-1}\left(1\right)=\frac{\mathrm{\pi }}{4}\end{array}$