Question

Find the value of \[ \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) \]

• A. \( \frac{1}{2} \cos^{-1}\left(\frac{3}{5}\right) \)
• B. \( \frac{1}{2} \sin^{-1}\left(\frac{3}{5}\right) \)
• C. \( \frac{1}{2} \tan^{-1}\left(\frac{3}{5}\right) \)
• D. \( \tan^{-1}\left(\frac{1}{2}\right) \)

Hint:

Use the addition formula for inverse tangents to simplify expressions involving the sum of two inverse tangents. This is particularly useful for expressions where the arguments are fractions.

Answer 1

The Correct Option is D

Step 1: We are given the expression \( \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) \). 
To simplify this, we use the addition formula for inverse tangents: 
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \] where \( a = \frac{1}{4} \) and \( b = \frac{2}{9} \). 
Step 2: Substitute the values of \( a \) and \( b \) into the formula: 
\[ \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = \tan^{-1}\left(\frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{1}{4} \cdot \frac{2}{9}}\right). \] 
Simplify the numerator: 
\[ \frac{1}{4} + \frac{2}{9} = \frac{9}{36} + \frac{8}{36} = \frac{17}{36}. \] 
Now simplify the denominator: 
\[ 1 - \frac{1}{4} \cdot \frac{2}{9} = 1 - \frac{2}{36} = 1 - \frac{1}{18} = \frac{17}{18}. \] 
So, the expression becomes: 
\[ \tan^{-1}\left(\frac{\frac{17}{36}}{\frac{17}{18}}\right) = \tan^{-1}\left(\frac{17}{36} \times \frac{18}{17}\right) = \tan^{-1}\left(\frac{18}{36}\right) = \tan^{-1}\left(\frac{1}{2}\right). \] 
Step 3: Thus, the value of \( \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) \) is \( \tan^{-1}\left(\frac{1}{2}\right) \).