Let M and m respectively be the maximum and minimum values of the function $f\left(x\right)=Ta{n}^{-1}(\sin x+\cos x)$  in  $[0,\frac{\pi }{2}]$  then the value of  $M+m=$

$f\left(x\right)=Ta{n}^{-1}(\sin x+\cos x)$  , where  $x\in [0,\frac{\pi }{2}]$
w.k.t.  $1\le \sin x+\cos x\le 2$
 $\Rightarrow Ta{n}^{-1}\left(1\right)\le f\left(x\right)\le Ta{n}^{-1}\sqrt{2}$
  $\therefore m=Ta{n}^{-1}\left(1\right) M=Ta{n}^{-1}\left(\sqrt{2}\right)$ 
 $M+m=Ta{n}^{-1}\sqrt{2}+Ta{n}^{-1}\left(1\right)$
  $=\pi +Ta{n}^{-1}\left(\frac{\sqrt{2}+1}{1-\sqrt{2}}\right) (\because xy>1)$ 
$=\pi +Ta{n}^{-1}\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right) = \pi -Ta{n}^{-1}(3+2\sqrt{2})$