JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
If \(\sum_{r=1}^{50} \tan ^{-1} \frac{1}{2 r^{2}}=p\), then the value of \(\tan p\) is :
- A \(\frac{101}{102}\)
- B \(\frac{51}{50}\)
- C \(100\)
- D \(\frac{50}{51}\)
Answer & Solution
Correct Answer
(D) \(\frac{50}{51}\)
Step-by-step Solution
Detailed explanation
\(\sum_{r=1}^{50} \tan ^{-1}\left(\frac{2}{4 r^{2}}\right)=\sum_{r=1}^{50} \tan ^{-1}\left(\frac{(2 r+1)-(2 r-1)}{1+(2 r+1)(2 r-1)}\right)\) \(\sum_{r=1}^{50} \tan ^{-1}(2 r+1)-\tan ^{-1}(2 r-1)\) \(\tan ^{-1}(101)-\tan ^{-1} 1 \Rightarrow \tan ^{-1} \frac{50}{51}\)
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