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JEE Mains · Maths · STD 12 - 2. inverse trigonometric function

The sum of the infinite series
\(\cot ^{-1}\left(\frac{7}{4}\right)+\cot ^{-1}\left(\frac{19}{4}\right)+\) \(\cot ^{-1}\left(\frac{39}{4}\right)+\cot ^{-1}\left(\frac{67}{4}\right)+\ldots .\) is :-

  1. A \(\frac{\pi}{2}+\tan ^{-1}\left(\frac{1}{2}\right)\)
  2. B \(\frac{\pi}{2}-\cot ^{-1}\left(\frac{1}{2}\right)\)
  3. C \(\frac{\pi}{2}+\cot ^{-1}\left(\frac{1}{2}\right)\)
  4. D \(\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{2}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{2}\right)\)

Step-by-step Solution

Detailed explanation

\begin{aligned} & \mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\frac{4}{4 \mathrm{n}^2+3}\right) \\ & \mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\frac{\left(\mathrm{n}+\frac{1}{2}\right)-\left(\mathrm{n}-\frac{1}{2}\right)}{1+\left(\mathrm{n}+\frac{1}{2}\right)\left(\mathrm{n}-\frac{1}{2…

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