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JEE Mains · Maths · STD 12 - 2. inverse trigonometric function

The value of \(\lim\limits _{n \rightarrow \infty} 6 \tan \left\{\sum\limits_{r=1}^{n} \tan ^{-1}\left(\frac{1}{r^{2}+3 r+3}\right)\right\}\) is equal to

  1. A \(1\)
  2. B \(2\)
  3. C \(3\)
  4. D \(6\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(3\)

Step-by-step Solution

Detailed explanation

\(T_{ r }=\tan ^{-1}\left[\frac{( r +2)-( r +1)}{1+( r +2)( r +1)}\right]\) \(=\tan ^{-1}( r +2)-\tan ^{-1}( r +1)\) \(T _{1}=\tan ^{-1} 3-\tan ^{-1} 2\) \(T _{2}=\tan ^{-1} 4-\tan ^{-1} 3\) \(T _{ n }=\tan ^{-1}( n +2)-\tan ^{-1}( n +1)\) _______________________________…