Question

\( \tan^{-1}\frac{3}{5} + \tan^{-1}\frac{6}{41} + \tan^{-1}\frac{9}{191} = \)

• A. \( \tan^{-1}\frac{9}{10} \)
• B. \( \tan^{-1}\frac{18}{19} \)
• C. \( \tan^{-1}\frac{3}{191} \)
• D. \( \tan^{-1}\frac{6}{205} \)

Hint:

Look for factors like 17, 13, 19 when simplifying large fractions in inverse trigonometry problems. Simplification at intermediate steps reduces calculation errors.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

We evaluate the sum of inverse tangent functions by grouping terms and using the addition formula iteratively.
Step 2: Key Formula or Approach:

\( \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \) for \( xy \textless 1 \).
Step 3: Detailed Explanation:

First Addition: \( \tan^{-1}\frac{3}{5} + \tan^{-1}\frac{6}{41} \) \[ = \tan^{-1}\left(\frac{\frac{3}{5} + \frac{6}{41}}{1 - \frac{3}{5}\cdot\frac{6}{41}}\right) = \tan^{-1}\left(\frac{\frac{123+30}{205}}{\frac{205-18}{205}}\right) \] \[ = \tan^{-1}\left(\frac{153}{187}\right) \] Simplify fraction: \( 153 = 9 \times 17 \), \( 187 = 11 \times 17 \). So, \( \tan^{-1}\left(\frac{9}{11}\right) \). Second Addition: \( \tan^{-1}\frac{9}{11} + \tan^{-1}\frac{9}{191} \) \[ = \tan^{-1}\left(\frac{\frac{9}{11} + \frac{9}{191}}{1 - \frac{9}{11}\cdot\frac{9}{191}}\right) \] Numerator: \( \frac{9(191) + 9(11)}{11 \cdot 191} = \frac{9(191+11)}{2101} = \frac{9(202)}{2101} \). Denominator: \( \frac{11(191) - 81}{2101} = \frac{2101 - 81}{2101} = \frac{2020}{2101} \). \[ = \tan^{-1}\left(\frac{9 \times 202}{2020}\right) \] Since \( 2020 = 10 \times 202 \): \[ = \tan^{-1}\left(\frac{9}{10}\right) \]
Step 4: Final Answer:

The sum is \( \tan^{-1}\frac{9}{10} \).