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Q.
The value of $5 \cot \left(\displaystyle\sum_{ k =1}^5\left(\cot ^{-1}\left( k ^2+ k +1\right)\right)\right)$ is equal to
Inverse Trigonometric Functions
Solution:
$\text { Consider } \left.\displaystyle\sum_{ k =1}^5\left(\tan ^{-1}\left(\frac{( k +1)- k }{1+ k ( k +1)}\right)\right)\right)=\tan ^{-1}( k +1)-\tan ^{-1} k $
$T _1=\tan ^{-1}(2)-\tan ^{-1}(1) ; T _2=\tan ^{-1}(3)-\tan ^{-1}(2) \text { and so on } $
$\text { hence } S =\tan ^{-1}(6)-\tan ^{-1}(1)=\tan ^{-1}\left(\frac{5}{7}\right)=\cot ^{-1}\left(\frac{7}{5}\right) $
$\therefore 5 \cot \left(\cot ^{-1} \frac{7}{5}\right)=7$