Trigonometry & Inverse Trigonometry
Telescoping sum of inverse tangents
nta_pyq_2023_jan
Grade 12
Question:
Let a_1 = 1, a_2, a_3, a_4, \ldots be consecutive natural numbers. Then \tan^{-1}\left(\frac{1}{1+a_1 a_2}\right) + \tan^{-1}\left(\frac{1}{1+a_2 a_3}\right) + \ldots + \tan^{-1}\left(\frac{1}{1+a_{2021}a_{2022}}\right) \text{ is equal to}
\frac{\pi}{4} - \cot^{-1}(2022)
\cot^{-1}(2022) - \frac{\pi}{4}
\tan^{-1}(2022) - \frac{\pi}{4}
\frac{\pi}{4} - \tan^{-1}(2022)
Step-by-Step Solution
Key Concept: Use telescoping: \tan^{-1}\left(\frac{a_{k+1}-a_k}{1+a_k a_{k+1}}\right) = \tan^{-1}(a_{k+1}) - \tan^{-1}(a_k). Sum telescopes to \tan^{-1}(2022) - \tan^{-1}(1).
Telescoping gives \tan^{-1}(2022) - \tan^{-1}(1) = \tan^{-1}(2022) - \pi/4. Also equals \pi/4 - \cot^{-1}(2022).
Correct Answer: 3