Question

The value of \(\tan^{-1}(\dfrac{1}{2})+\tan^{-1}(\dfrac{2}{5})\) is ? 

• A. \(\tan^{-1} (5)\)
• B. \(\tan^{-1}(\dfrac{1}{5})\)
• C. \(\tan^{-1} (\dfrac{2}{3})\)
• D. \(\tan^{-1} (\dfrac{8}{9})\)
• E. \(\tan^{-1} (\dfrac{9}{8})\)

Answer 1

Answer: (E)

Step 1: Apply the tangent addition formula for inverse functions: \[ \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left( \frac{a + b}{1 - ab} \right) \quad \text{if} \quad ab < 1 \] Here, \( a = \frac{1}{2} \) and \( b = \frac{2}{5} \), so \( ab = \frac{1}{5} < 1 \).

Step 2: Substitute the values: \[ \tan^{-1}\left( \frac{1}{2} \right) + \tan^{-1}\left( \frac{2}{5} \right) = \tan^{-1}\left( \frac{\frac{1}{2} + \frac{2}{5}}{1 - \frac{1}{2} \cdot \frac{2}{5}} \right) \]

Step 3: Simplify the expression: \[ = \tan^{-1}\left( \frac{\frac{5}{10} + \frac{4}{10}}{1 - \frac{2}{10}} \right) = \tan^{-1}\left( \frac{\frac{9}{10}}{\frac{8}{10}} \right) = \tan^{-1}\left( \frac{9}{8} \right) \]

Conclusion: The exact value is \( \tan^{-1}\left( \frac{9}{8} \right) \).

Answer 2

Let \( A = \tan^{-1}\left(\frac{1}{2}\right) \) and \( B = \tan^{-1}\left(\frac{2}{5}\right) \). We want to find \( A + B \).

We can use the tangent addition formula:

\[ \tan(A + B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)} \]

Substituting the given values:

\[ \tan(A + B) = \frac{\left(\frac{1}{2}\right) + \left(\frac{2}{5}\right)}{1 - \left(\frac{1}{2}\right)\left(\frac{2}{5}\right)} = \frac{\frac{5 + 4}{10}}{1 - \frac{1}{5}} = \frac{\frac{9}{10}}{\frac{4}{5}} = \frac{9}{10} \cdot \frac{5}{4} = \frac{9}{8} \]

Therefore, \( A + B = \tan^{-1}\left(\frac{9}{8}\right) \).