(i) Using the identity \(\tan(\pi - x) = -\tan x\), we have:

\(\tan(\pi - x) = -k\).

(ii) Using the identity \(\tan\left(\frac{\pi}{2} - x\right) = \cot x = \frac{1}{\tan x}\), we have:

\(\tan\left(\frac{\pi}{2} - x\right) = \frac{1}{k}\).

(iii) Using the identity \(\sin x = \frac{\tan x}{\sqrt{1 + \tan^2 x}}\), we have:

\(\sin x = \frac{k}{\sqrt{1 + k^2}}\).