Step 1: Understanding the Concept:
Inverse trigonometric functions like \(\tan^{-1}(x)\) represent angles. Adding two inverse tangent terms is equivalent to finding the angle resulting from the sum of two other angles. To perform this operation, we utilize the inverse tangent addition identity, which is derived from the \(\tan(A+B)\) formula from basic trigonometry.
However, we must be careful with the range. The standard addition formula is valid only under specific conditions related to the product of the two arguments (\(xy\)). If the product exceeds \(1\), the resulting angle may fall outside the principal branch \((-\pi/2, \pi/2)\) and require an adjustment of \(\pi\).
Key Formula or Approach:
The addition identity for inverse tangents is:
\[ \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \]
This specific form is used when \(xy<1\).
Step 2: Detailed Explanation:
Let \(x = 1/2\) and \(y = 1/3\).
First, check the condition for the product:
\[ xy = \left(\frac{1}{2}\right)\left(\frac{1}{3}\right) = \frac{1}{6} \]
Since \(1/6<1\), the direct application of the formula is valid.
Substitute the values into the identity:
\[ \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left( \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \cdot \frac{1}{3}} \right) \]
Simplify the numerator:
\[ \frac{1}{2} + \frac{1}{3} = \frac{3+2}{6} = \frac{5}{6} \]
Simplify the denominator:
\[ 1 - \frac{1}{6} = \frac{5}{6} \]
Now, calculate the final ratio inside the function:
\[ \text{Expression} = \tan^{-1}\left( \frac{5/6}{5/6} \right) = \tan^{-1}(1) \]
The principal value of \(\tan^{-1}(1)\) is the angle \(\theta\) in the interval \((-\pi/2, \pi/2)\) such that \(\tan\theta = 1\). This angle is \(\pi/4\) (or \(45^{\circ}\)).
Step 3: Final Answer:
The sum of \(\tan^{-1}(1/2)\) and \(\tan^{-1}(1/3)\) simplifies to \(\tan^{-1}(1)\), which has a principal value of \(\pi/4\).