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Q. If $S _{ n }=\displaystyle\sum_{ r =1}^{ n } r$ ! then for $n >6 \quad\left(\right.$ given $\left.\displaystyle\sum_{ r =1}^6 r !=873\right)$
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)

Inverse Trigonometric Functions

Solution:

$S _{ n }=\underline{1 !+2 !+3 !+4 !+5 !+6 !}+71$ where I be an integer
$S _{ n }=873+7 I $
$\frac{ S _{ n }}{7}=124.71+ I \Rightarrow \left[\frac{ S _{ n }}{7}\right]=124+ I$
$\therefore 7\left[\frac{ S _{ n }}{7}\right]=868+7 I \Rightarrow S _{ n }-7\left[\frac{ S _{ n }}{7}\right]=(873+7 I )-(868+7 I )=5 $
$\text { now } \sin ^{-1}(\sin 5)=\sin ^{-1}(\sin (5-2 \pi))=5-2 \pi \Rightarrow \text { (P) } $
$\cos ^{-1}(\cos 5)=\cos ^{-1}(\cos (2 \pi-5))=2 \pi-5 \Rightarrow \text { (Q) } $
$\tan ^{-1}(\tan 5)=\tan ^{-1}(\tan (5-2 \pi))=5-2 \pi \Rightarrow ( P ) $
$\text { and } \cot ^{-1}(\cot 5)=\cot ^{-1}(\cot (5-\pi))=5-\pi \Rightarrow \text { (R) } $