If  $S_n=\sum _{r=1}^{n}r!thenforn>6\left(given\sum _{r=1}^{6}r!=873\right)$

	Column – I		Column – II
A)	${\sin }^{-1}\left(\sin \left(S_n-7\left[\frac{S_n}{7}\right]\right)\right)$	P)	$5-2\pi$
B)	${\cos }^{-1}\left(\cos \left(S_n-7\left[\frac{S_n}{7}\right]\right)\right)$	Q)	$2\pi -5$
C)	${\tan }^{-1}\left(\tan \left(S_n-7\left[\frac{S_n}{7}\right]\right)\right)$	R)	$6-2\pi$
D)	${\cot }^{-1}\left(\cot \left(S_n-7\left[\frac{S_n}{7}\right]\right)\right)$	S)	$5-\pi$
		T)	$\pi -4$

(where [ ] denotes greatest integer function)  
The correct match :

$S_n=1!+2!+3!+4!+5!+6!+7I$  where I be an integer. 
 $S_n=873+7I$
 $\frac{S_n}{7}=124.71+I\Rightarrow \left[\frac{S_n}{7}\right]=124+I$
 $\therefore 7\left[\frac{S_n}{7}\right]=868+7I\Rightarrow S_n-7\left[\frac{S_n}{7}\right]=\left(873+7I\right)-\left(868+7I\right)=5$
Now,  ${\sin }^{-1}\left(\sin 5\right)={\sin }^{-1}\left(\sin \left(5-2\pi \right)\right)=5-2\pi \Rightarrow \left(P\right)$
 $$\begin{gathered} {\cos }^{-1}\left(\cos 5\right)={\cos }^{-1}\left(\cos \left(2\pi -5\right)\right)=2\pi -5\Rightarrow \left(Q\right) \\ {\tan }^{-1}\left(\tan 5\right)={\tan }^{-1}\left(\tan \left(5-2\pi \right)\right)=5-2\pi \Rightarrow \left(P\right) \\ and{\cot }^{-1}\left(\cot 5\right)=co{t}^{-1}\left(\cot \left(5-\pi \right)\right)=5-\pi \Rightarrow \left(S\right) \end{gathered}$$