Question

If $ \theta = \tan^{-1} \left( \frac{1}{3} \right) + \tan^{-1} \left( \frac{1}{7} \right) + \tan^{-1} \left( \frac{1}{13} \right) + \tan^{-1} \left( \frac{1}{21} \right) + \tan^{-1} \left( \frac{1}{31} \right) $, then $ \tan \theta = ? $

• A. \( \frac{3}{5} \)
• B. 1
• C. \( \frac{5}{7} \)
• D. \( \frac{7}{9} \)

Hint:

To sum inverse tangents, apply \( \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \) when \( ab<1 \).

Answer 1

The Correct Option is C

Use identity: \[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right),\text{ when } ab<1 \] Group and simplify: Step 1: \[ \tan^{-1} \left( \frac{1}{3} \right) + \tan^{-1} \left( \frac{1}{7} \right) = \tan^{-1} \left( \frac{\frac{1}{3} + \frac{1}{7}}{1 - \frac{1}{3} \cdot \frac{1}{7}} \right) = \tan^{-1} \left( \frac{\frac{10}{21}}{1 - \frac{1}{21}} \right) = \tan^{-1} \left( \frac{10}{20} \right) = \tan^{-1} \left( \frac{1}{2} \right) \] Repeat and finally reduce: All simplify to: \[ \tan \theta = \frac{5}{7} \]