If tan⁻¹k+tan⁻¹ \(\frac{1}{k}\)=\(\frac{π}{2} \), then what is the value of k?

Calculation:

Given,

\( \tan^{-1}(k) + \tan^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{4} \)

Use the addition formula for inverse tangents:

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

Using this for \(\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{2}\right) \), we get:

\( \tan^{-1}\left(\frac{k + \frac{1}{2}}{1 - k \cdot \frac{1}{2}}\right) = \frac{\pi}{4} \)

Since \(\tan\left(\frac{\pi}{4}\right) = 1 \), we have:

\( \frac{k + \frac{1}{2}}{1 - \frac{k}{2}} = 1 \)

\( k + \frac{1}{2} = 1 - \frac{k}{2} \)

\( 2k + 1 = 2 - k \)

\( 2k + k = 2 - 1 \)

\( 3k = 1 \)

\( k = \frac{1}{3} \)

Hence, the correct answer is Option 3.