126. Show that the sum arctan(x)+arctan(1/x) is constant.

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Recall the formula for the tangent of a sum: \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\). This will help us analyze the sum \(\arctan(x) + \arctan(1/x)\) by considering its tangent.

Let \(A = \arctan(x)\) and \(B = \arctan(1/x)\). Then, \(\tan A = x\) and \(\tan B = \frac{1}{x}\) (assuming \(x \neq 0\)). Substitute these into the tangent sum formula:

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

Notice that the denominator \(1 - 1 = 0\), which means \(\tan(A + B)\) is undefined (tends to infinity). This implies that \(A + B\) corresponds to an angle where the tangent function has a vertical asymptote, specifically \(\frac{\pi}{2}\) or \(-\frac{\pi}{2}\) depending on the sign of \(x\).

Therefore, conclude that \(\arctan(x) + \arctan(1/x)\) is constant and equal to \(\frac{\pi}{2}\) (or \(-\frac{\pi}{2}\)) for all \(x \neq 0\), showing the sum is constant.

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Properties of the Arctangent Function

The arctangent function, denoted arctan(x), is the inverse of the tangent function restricted to its principal domain. It returns an angle whose tangent is x, typically in the interval (-π/2, π/2). Understanding its behavior and range is essential for manipulating expressions involving arctan.

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The sum of inverse trigonometric functions like arctan(x) + arctan(y) can often be simplified using angle addition formulas. Specifically, arctan(a) + arctan(b) = arctan((a + b)/(1 - ab)) when the expression is defined, which helps in proving constancy or evaluating sums.

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When dealing with expressions like arctan(x) + arctan(1/x), it is important to consider the domain of x (excluding zero) and the continuity of the function. This ensures the sum is well-defined and helps in showing that the sum remains constant for all valid x.

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