Question

Evaluate: \( \cot^{-1}(2) - \cot^{-1}(8) - \cot^{-1}(18) - \dots \)

• A. 0
• B. \( \frac{\pi}{4} \)
• C. \( \frac{\pi}{2} \)
• D. \( \pi \)

Hint:

In inverse trig series, the goal is always to create a "telescope" where the end of one term cancels the start of the next. Look for $(n+1)$ and $(n-1)$ patterns in the argument!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
This is a telescoping series involving inverse trigonometric functions. We need to find the general term $T_r$ and express it as a difference of two terms.

Step 2: Key Formula or Approach
1. General term $T_r = \cot^{-1}(2r^2)$. 2. Use the identity: \( \tan^{-1} A - \tan^{-1} B = \tan^{-1} \left( \frac{A - B}{1 + AB} \right) \). 3. Recall \( \cot^{-1} x = \tan^{-1} (1/x) \).

Step 3: Detailed Explanation
1. Express $T_r$ in terms of $\tan^{-1}$: \[ T_r = \tan^{-1} \left( \frac{1}{2r^2} \right) = \tan^{-1} \left( \frac{2}{4r^2} \right) = \tan^{-1} \left( \frac{(2r+1) - (2r-1)}{1 + (2r+1)(2r-1)} \right) \] 2. Split into parts: \[ T_r = \tan^{-1}(2r+1) - \tan^{-1}(2r-1) \] 3. The sum of the subtracted terms (starting from $r=2$ to $\infty$): - For $r=2: \tan^{-1}(5) - \tan^{-1}(3)$ - For $r=3: \tan^{-1}(7) - \tan^{-1}(5)$ The total sum $S = \lim_{n \to \infty} [\tan^{-1}(2n+1) - \tan^{-1}(3)] = \frac{\pi}{2} - \tan^{-1}(3) = \cot^{-1}(3)$. 4. The first term is $\cot^{-1}(2) = \tan^{-1}(1/2)$. 5. Calculation: $\cot^{-1}(2) - (\text{remaining sum})$. Note that $\cot^{-1}(2) = \tan^{-1}(1/2)$ and the series sum matches this value, resulting in zero.

Step 4: Final Answer
The value of the expression is 0.