TS EAMCET · Maths · Inverse Trigonometric Functions
If \(\sum_{n=1}^k \tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right)=\tan ^{-1} \alpha\), then \(\alpha=\)
- A \(\frac{k}{k+2}\)
- B \(\frac{2 k}{2 k+1}\)
- C \(\frac{k}{2 k+5}\)
- D \(\frac{3 k}{4 k+5}\)
Answer & Solution
Correct Answer
(C) \(\frac{k}{2 k+5}\)
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