Question

\( \tan^{-1} 1/2 + \tan^{-1} 1/3 = \tan^{-1} 1/5 \).

Hint:

For the sum of inverse tangents, use the addition formula: \( \tan^{-1} a + \tan^{-1} b = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \).

Answer 1

Step 1: Applying the addition formula for inverse tangents.
We can use the formula for the sum of inverse tangents: \[ \tan^{-1} a + \tan^{-1} b = \tan^{-1}\left(\frac{a + b}{1 - ab}\right), \] if \( ab < 1 \). Substituting \( a = \frac{1}{2} \) and \( b = \frac{1}{3} \): \[ \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3} = \tan^{-1}\left(\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \times \frac{1}{3}}\right). \]

Step 2: Simplifying the expression.
The numerator is: \[ \frac{1}{2} + \frac{1}{3} = \frac{3 + 2}{6} = \frac{5}{6}. \] The denominator is: \[ 1 - \frac{1}{2} \times \frac{1}{3} = 1 - \frac{1}{6} = \frac{5}{6}. \] Thus: \[ \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3} = \tan^{-1} \left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1} 1. \] Since \( \tan^{-1} 1 = \frac{\pi}{4} \), the equation is false.

Step 3: Conclusion.
Thus, \( \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3} \neq \tan^{-1} \frac{1}{5} \), making the statement false.