If k is a root of $x{}^2-4x+1=0$, then what is tan⁻¹k+tan⁻¹\(\frac{1}{k}\) equal to?

Calculation:

We are given that k is a root of $x{}^2-4x+1=0$

Solving the quadratic equation:

\(x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} \)

⇒ \(x = 2 \pm \sqrt{3} \)

Thus, the two possible values of k are

⇒ \(k = 2 + \sqrt{3} \quad \text{or} \quad k = 2 - \sqrt{3} \)

We need to find  \(\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right) \)

Using the identity for the sum of inverse tangents

\(\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \)

⇒ \(\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right) = \tan^{-1}\left(\frac{k + \frac{1}{k}}{1 - k \cdot \frac{1}{k}}\right) \)

\(= \tan^{-1}\left(\frac{k + \frac{1}{k}}{1 - 1}\right) = \tan^{-1} \left(\frac{k + \frac{1}{k}}{0}\right) \)

This expression results in an undefined value, but we know from properties of the inverse tangent that

⇒ \(\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right) = \frac{\pi}{2}\)

Hence, the correct answer is Option 4.