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
- A \(1\)
- B \(2\)
- C \(3\)
- D \(6\)
Answer & Solution
Correct Answer
(C) \(3\)
Step-by-step Solution
Detailed explanation
\(T_{ r }=\tan ^{-1}\left[\frac{( r +2)-( r +1)}{1+( r +2)( r +1)}\right]\) \(=\tan ^{-1}( r +2)-\tan ^{-1}( r +1)\) \(T _{1}=\tan ^{-1} 3-\tan ^{-1} 2\) \(T _{2}=\tan ^{-1} 4-\tan ^{-1} 3\) \(T _{ n }=\tan ^{-1}( n +2)-\tan ^{-1}( n +1)\) _______________________________…
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