Let f(x)=sin x +cos x+ tan x+ arc sin x+ arc cos x + arc tan x . If M and m are maximum and minimum values of f(x) , then their arithmetic mean is equal to 

Domain of f is [-1,1]

$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sinx}+\mathrm{cosx}+\mathrm{tanx}+{\sin }^{-1}\mathrm{x}+{\cos }^{-1}\mathrm{x}+{\tan }^{-1}\mathrm{x}$

$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sinx}+\mathrm{cosx}+\mathrm{tanx}+\frac{\mathrm{\pi }}{2}+{\tan }^{-1}\mathrm{x}$

$\mathrm{f}\text{'}\left(\mathrm{x}\right)=\mathrm{cosx}-\mathrm{sinx}+{\sec }^2\mathrm{x}+0+\frac{1}{1+{\mathrm{x}}^2}$ 

here,${\sec }^2\ge 1$ and $\frac{1}{1+{\mathrm{x}}^2}\in \left[\frac{1}{2},1\right]$ and  $\mathrm{cosx}-\mathrm{sinx} \in \left[-\sqrt{2},\sqrt{2}\right]$

hence $\mathrm{f}\text{'}\left(\mathrm{x}\right)>0\Rightarrow$ f is increasing

Range is [f(-1),f(1)]

$\therefore {\left[\mathrm{f}\left(\mathrm{x}\right)\right]}_{\min }=\mathrm{f}\left(-1\right)=-\sin 1+\cos 1-\tan 1+\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{4}$

                   $m=\frac{\mathrm{\pi }}{4}+\cos 1-\sin 1-\tan 1$

$\therefore {\left[\mathrm{f}\left(\mathrm{x}\right)\right]}_{max}=\mathrm{f}\left(1\right)=\sin 1+\cos 1+\tan 1+\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{4}$

                      $M=\frac{3\mathrm{\pi }}{4}+\cos 1+\sin 1+\tan 1$

                       $\Rightarrow \frac{\mathrm{M}+\mathrm{m}}{2}=\frac{\mathrm{\pi }}{2}+\cos 1$