Solving Arctan(k) + Arctan(1/k) Given K Is A Root Of X^2 - 4x + 1 = 0

In the realm of trigonometry and inverse trigonometric functions, there exist numerous fascinating relationships and identities. One such intriguing problem involves finding the value of an expression containing inverse tangents. Specifically, we aim to determine the value of ${ \tan^{-1}k + \tan^{-1}\frac{1}{k} }$, given that ${k}$ is a root of the quadratic equation ${ x^2 - 4x + 1 = 0 }$. This exploration will involve employing our understanding of quadratic equations, their roots, and the properties of inverse trigonometric functions. The final answer reveals a beautiful interplay between algebra and trigonometry, highlighting how seemingly disparate mathematical concepts are intricately connected. Let's dive into the problem and unravel the solution step-by-step, ensuring each concept is thoroughly understood.

The quadratic equation presented is ${ x^2 - 4x + 1 = 0 }$. To understand the nature of the root k, we can employ the quadratic formula, which states that for an equation of the form ${ ax^2 + bx + c = 0 }$, the roots are given by: ${ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} }$ In our case, ${ a = 1 }$, ${ b = -4 }$, and ${ c = 1 }$. Substituting these values into the quadratic formula, we get: ${ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} }$ Thus, the two roots of the equation are ${ 2 + \sqrt{3} }$ and ${ 2 - \sqrt{3} }$. Let's denote these roots as ${ k_1 = 2 + \sqrt{3} }$ and ${ k_2 = 2 - \sqrt{3} }$. Since ${k}$ is a root of the equation, it can be either ${ k_1 }$ or ${ k_2 }$. An important observation here is that ${ k_1 }$ and ${ k_2 }$ are reciprocals of each other. Specifically: ${ \frac{1}{k_1} = \frac{1}{2 + \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3} = k_2 }$ This reciprocal relationship between the roots will be crucial in simplifying the expression involving inverse tangents. The roots are real and positive, which helps in determining the range of the inverse tangent function. The next step involves utilizing these roots to evaluate the given expression ${ \tan^{-1}k + \tan^{-1}\frac{1}{k} }$. By understanding the nature and relationship between the roots of the quadratic equation, we lay the groundwork for solving the problem using trigonometric identities.

The inverse tangent function, denoted as ${ \tan^{-1}(x) }$ or ${ \arctan(x) }$, is the inverse of the tangent function. It returns the angle whose tangent is ${x}$. The principal value range of the inverse tangent function is ${ (-\frac{\pi}{2}, \frac{\pi}{2}) }$. A crucial property of the inverse tangent function that we will use is: ${ \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \begin{cases} \frac{\pi}{2}, & \text{if } x > 0 \\ -\frac{\pi}{2}, & \text{if } x < 0 \end{cases} }$ This property arises from the complementary relationship between angles in a right-angled triangle. If ${ \theta = \tan^{-1}(x) }$, then ${ \tan(\theta) = x }$. Consider a right-angled triangle where the opposite side is ${x}$ and the adjacent side is 1. The angle ${ \theta }$ is such that its tangent is ${x}$. Now, if we consider the reciprocal, ${ \frac{1}{x} }$, this corresponds to the tangent of the complementary angle, which is ${ \frac{\pi}{2} - \theta }$. Thus, ${ \tan(\frac{\pi}{2} - \theta) = \frac{1}{x} }$, and ${ \tan^{-1}(\frac{1}{x}) = \frac{\pi}{2} - \theta }$. Adding ${ \tan^{-1}(x) }$ and ${ \tan^{-1}(\frac{1}{x}) }$ yields: ${ \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \theta + \left(\frac{\pi}{2} - \theta\right) = \frac{\pi}{2} }$ This holds true when ${x}$ is positive. If ${x}$ is negative, a similar argument can be made, considering the range of the inverse tangent function, which leads to the result ${ -\frac{\pi}{2} }$. The understanding of this property is pivotal in simplifying the given expression. In our case, we have already established that the roots ${k}$ of the quadratic equation are positive. Therefore, we can directly apply the property for positive ${x}$ values. This trigonometric identity allows us to bypass complex calculations and arrive at a straightforward solution. The next section will demonstrate how this property is applied to solve the problem at hand.

Given that ${k}$ is a root of the equation ${ x^2 - 4x + 1 = 0 }$, we found that the roots are ${ k = 2 \pm \sqrt{3} }$. We also observed that the roots are reciprocals of each other, meaning if ${ k = 2 + \sqrt{3} }$, then ${ \frac{1}{k} = 2 - \sqrt{3} }$, and vice versa. Furthermore, both roots are positive, as ${ 2 > \sqrt{3} }$.

Now, we need to evaluate ${ \tan^{-1}k + \tan^{-1}\frac{1}{k} }$. We can use the property of inverse tangent functions that states: ${ \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \begin{cases} \frac{\pi}{2}, & \text{if } x > 0 \\ -\frac{\pi}{2}, & \text{if } x < 0 \end{cases} }$ Since ${k}$ is a root of the given quadratic equation, and we have established that both roots are positive (${ 2 + \sqrt{3} > 0 }$ and ${ 2 - \sqrt{3} > 0 }$), we can apply the first case of the property. Therefore: ${ \tan^{-1}k + \tan^{-1}\frac{1}{k} = \frac{\pi}{2} }$ This result holds true regardless of which root we choose for ${k}$, because the property is applicable for any positive value of ${x}$. This elegant solution demonstrates the power of using trigonometric identities to simplify expressions. By recognizing the reciprocal relationship and applying the inverse tangent property, we avoid the need for complicated calculations involving the actual values of the roots. The final answer, ${ \frac{\pi}{2} }$, highlights the inherent symmetry and mathematical structure within the problem. Thus, the value of ${ \tan^{-1}k + \tan^{-1}\frac{1}{k} }$ is ${ \frac{\pi}{2} }$.

In conclusion, the problem of finding the value of ${ \tan^{-1}k + \tan^{-1}\frac{1}{k} }$, where ${k}$ is a root of the equation ${ x^2 - 4x + 1 = 0 }$, elegantly showcases the interplay between algebra and trigonometry. By first solving the quadratic equation, we identified the roots and their reciprocal relationship. We then leveraged a key property of the inverse tangent function to simplify the expression. Specifically, we used the property that ${ \tan^{-1}(x) + \tan^{-1}(\frac{1}{x}) = \frac{\pi}{2} }$ for ${ x > 0 }$. Since both roots of the quadratic equation were found to be positive, we directly applied this property to obtain the result ${ \frac{\pi}{2} }$. This problem underscores the importance of recognizing fundamental properties and relationships in mathematics. The solution path involves not only algebraic manipulation but also a sound understanding of trigonometric functions and their inverses. The problem’s elegance lies in its simplicity, achieved through the appropriate application of mathematical principles. The final answer, ${ \frac{\pi}{2} }$, is a testament to the inherent beauty and harmony within mathematical structures. This exercise serves as a valuable illustration of how diverse mathematical concepts can coalesce to provide concise and meaningful solutions, reinforcing the interconnected nature of mathematical knowledge and problem-solving techniques. The ability to identify and apply relevant properties and theorems is a crucial skill in mathematics, and this example perfectly demonstrates its effectiveness.