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JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
\(S = {\tan ^{ - 1}}\left( {\frac{1}{{{n^2} + n + 1}}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{{{n^2} + 3n + 3}}} \right) + ..... + {\tan ^{ - 1}}\left( {\frac{1}{{1 + \left( {n + 19} \right)\left( {n + 20} \right)}}} \right)\) , then \(tan\,S\) is equal to
- A \(\frac{{20}}{{401 + 20n}}\)
- B \(\frac{n}{{{n^2} + 20n + 1}}\)
- C \(\frac{20}{{{n^2} + 20n + 1}}\)
- D \(\frac{n}{{401 + 20n}}\)
Answer & Solution
Correct Answer
(C) \(\frac{20}{{{n^2} + 20n + 1}}\)
Step-by-step Solution
Detailed explanation
we know that, \({\tan ^{ - 1}}\frac{1}{{1 + 2}} + {\tan ^{ - 1}}\frac{1}{{1 + 2 \times 3}} + {\tan ^{ - 1}}\frac{1}{{1 + 3 \times 4}} + ... + \) \({\tan ^{ - 1}}\frac{1}{{1 + \left( {n - 1} \right)n}} + {\tan ^{ - 1}}\frac{1}{{1 + n\left( {n + 1} \right)}} + .... + \)…
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