Let ‘A’ be the domain of $\frac{1}{\sqrt{-{\mathrm{x}}^2+3\mathrm{x}}}$. If $\lambda$ is the length of the set ‘A’ then${\mathrm{Tan}}^{-1}\left(\frac{1}{\mathrm{\lambda }-1}\right)+{\mathrm{Tan}}^{-1}\left(\frac{1}{\mathrm{\lambda }}\right)=$

Domain is given by
 $-{\mathrm{x}}^2+3\mathrm{x}>0\Rightarrow \mathrm{x}\in (0,3)$
$\therefore \mathrm{A}=(0,3),\text{ length of ' }\mathrm{A}\text{ ' is }3-0=3=\mathrm{\lambda }$
${\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}={\tan }^{-1}\left(\frac{{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}}{1-{\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{3}}}\right)$
$={\tan }^{-1}\left(\frac{{\displaystyle \frac{{\displaystyle 5}}{6}}}{{\displaystyle \frac{5}{6}}}\right)={\tan }^{-1}\left(1\right)=\frac{\mathrm{\pi }}{4}$