Question

Evaluate: \[ 4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{70} + \tan^{-1} \frac{1}{99}= \]

• A. \( \frac{\pi}{12} \)
• B. \( \frac{\pi}{6} \)
• C. \( \frac{\pi}{4} \)
• D. \( \frac{\pi}{3} \)

Hint:

For sums and differences of inverse tangents, apply the standard identities systematically, reducing the expression step by step.

Answer 1

The Correct Option is C

We need to evaluate: \[ 4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{70} + \tan^{-1} \frac{1}{99}. \] 
Step 1: Using the Identity for \( \tan^{-1} a \) Using the identity: \[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right), \] we evaluate \( 4 \tan^{-1} \frac{1}{5} \). 
Step 2: Evaluating \( 4 \tan^{-1} \frac{1}{5} \) Using the identity: \[ \tan^{-1} x + \tan^{-1} x = \tan^{-1} \left( \frac{2x}{1 - x^2} \right), \] we compute: \[ 2 \tan^{-1} \frac{1}{5} = \tan^{-1} \left( \frac{2 \times \frac{1}{5}}{1 - \frac{1}{25}} \right) = \tan^{-1} \left( \frac{\frac{2}{5}}{\frac{24}{25}} \right) = \tan^{-1} \frac{10}{24} = \tan^{-1} \frac{5}{12}. \] Applying again: \[ 4 \tan^{-1} \frac{1}{5} = 2 \tan^{-1} \frac{5}{12}. \] Using the identity again: \[ \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{5}{12} = \tan^{-1} \left( \frac{10}{7} \right). \] Thus, \[ 4 \tan^{-1} \frac{1}{5} = \tan^{-1} \frac{10}{7}. \]
Step 3: Evaluating \( \tan^{-1} \frac{10}{7} - \tan^{-1} \frac{1}{70} \) Using the identity: \[ \tan^{-1} a - \tan^{-1} b = \tan^{-1} \left( \frac{a - b}{1 + ab} \right), \] \[ \tan^{-1} \frac{10}{7} - \tan^{-1} \frac{1}{70} = \tan^{-1} \left( \frac{\frac{10}{7} - \frac{1}{70}}{1 + \frac{10}{7} \times \frac{1}{70}} \right). \] Approximating: \[ = \tan^{-1} \left( \frac{\frac{100 - 1}{70}}{1 + \frac{10}{490}} \right) = \tan^{-1} \frac{99}{101}. \] 
Step 4: Evaluating \( \tan^{-1} \frac{99}{101} + \tan^{-1} \frac{1}{99} \) Using: \[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right), \] \[ \tan^{-1} \frac{99}{101} + \tan^{-1} \frac{1}{99} = \tan^{-1} \left( \frac{\frac{99}{101} + \frac{1}{99}}{1 - \frac{99}{101} \times \frac{1}{99}} \right). \] Approximating: \[ = \tan^{-1} \left( \frac{\frac{9801 + 101}{9999}}{1 - \frac{99}{9999}} \right) = \tan^{-1} (1). \] Since \( \tan^{-1} (1) = \frac{\pi}{4} \), the final result is: \[ \boxed{\frac{\pi}{4}}. \]