Trigonometry & Inverse Trigonometry
Trigonometry
star_batch_jee_advanced_2025
Grade None
Question:
The value of $\cot^{-1}\left(2^2-\frac{1}{2}\right)+\cot^{-1}\left(2^3+\frac{1}{2^2}\right)+\cot^{-1}\left(2^4+\frac{1}{2^3}\right)+\ldots \infty$ is:
$\tan^{-1}\frac{1}{1}$
$\tan^{-1}\frac{1}{2}$
$1$
None of these
Step-by-Step Solution
Key Concept: The arctangent difference formula $\tan^{-1}a - \tan^{-1}b = \tan^{-1}\frac{a-b}{1+ab}$ creates a telescoping series.
We have $T_n = \cot^{-1}\left(2^{n+1} + \frac{1}{2^n}\right) = \tan^{-1}\frac{2^n(2-1)}{1+2^{n+1}\cdot 2^n} = \tan^{-1}2^{n+1} - \tan^{-1}2^n$. Summing from $n=1$ to $n$: $S_n = \tan^{-1}2^{n+1} - \tan^{-1}2$. Taking limit as $n \to \infty$: $S_\infty = \frac{\pi}{2} - \tan^{-1}2 = \cot^{-1}2 = \tan^{-1}\frac{1}{2}$.
Correct Answer: 2