Step 1: Rewrite
Divide numerator and denominator by $\cos x$: $f=\dfrac{k\tan x+2}{\tan x+1}$, a function of $t=\tan x$.
Step 2: Derivative in t
$\dfrac{df}{dt}=\dfrac{k(t+1)-(kt+2)}{(t+1)^2}=\dfrac{k-2}{(t+1)^2}$, and $\dfrac{dt}{dx}=\sec^2x\gt0$.
Step 3: Conclusion
The sign of $f'$ is the sign of $k-2$, so strict increase needs $k\gt2$.
Final Answer:
The condition is k greater than 2.
\[ \boxed{\text{(D) }k \gt 2} \]