Calculate the Sum of Tan Inverse Values

We are asked to find the value of the expression \({\tan ^{ - 1}}\left( {\frac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{3}} \right)\). This problem involves calculating the sum of two tan inverse functions.

Applying the Tan Inverse Sum Formula

To solve this, we use the standard sum formula for inverse tangent functions:

\({\tan ^{ - 1}}(x) + {\tan ^{ - 1}}(y) = {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right)\), provided \(xy < 1\).

This sum formula is crucial for simplifying expressions involving the sum of two tan inverse or arctan terms. It is a key identity in inverse trigonometric functions.

Step-by-Step Calculation

In our problem, we have \(x = \frac{1}{2}\) and \(y = \frac{1}{3}\).

First, let's check the condition \(xy < 1\):

\(xy = \left( {\frac{1}{2}} \right) \times \left( {\frac{1}{3}} \right) = \frac{1}{6}\).

Since \(\frac{1}{6} < 1\), the formula is applicable for these inverse trigonometric functions.

Now, substitute the values into the sum formula:

\({\tan ^{ - 1}}\left( {\frac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{3}} \right) = {\tan ^{ - 1}}\left( {\frac{{\frac{1}{2} + \frac{1}{3}}}{{1 - \left( {\frac{1}{2}} \right)\left( {\frac{1}{3}} \right)}}} \right)\)

Simplify the numerator of the fraction inside the tan inverse:

\(\frac{1}{2} + \frac{1}{3} = \frac{3 \times 1 + 2 \times 1}{6} = \frac{3 + 2}{6} = \frac{5}{6}\)

Simplify the denominator of the fraction inside the tan inverse:

\(1 - \left( {\frac{1}{2}} \right)\left( {\frac{1}{3}} \right) = 1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{6 - 1}{6} = \frac{5}{6}\)

Substitute these simplified values back into the tan inverse expression:

\({\tan ^{ - 1}}\left( {\frac{{\frac{5}{6}}}{{\frac{5}{6}}}} \right) = {\tan ^{ - 1}}(1)\)

Finding the Final Value of Tan Inverse(1)

The value of \({\tan ^{ - 1}}(1)\) is the angle \(\theta\) such that \(\tan(\theta) = 1\). In radians, this angle is \(\frac{\pi}{4}\). This is a standard value from basic trigonometry.

Therefore, the value of \({\tan ^{ - 1}}\left( {\frac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{3}} \right)\) is \(\frac{\pi}{4}\).

Conclusion

Using the tan inverse sum formula, we found that the value of \({\tan ^{ - 1}}\left( {\frac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{3}} \right)\) is \(\frac{\pi}{4}\).