ft,
Each time, we're talking about what are essentially four vectors in the vertical plane: W, L, T and D.
So let's not fully neglect the fact that they are not necessarily at right angles all the time.
In steady vertical flight, L and the normal component of T are obviously second-order. So, I garee with you, to explain the basic state of affairs, T=W+D makes perfect sense.
But to get back to the original question, I'm still convinced that in a "typical" aircraft in a "typical" steady climb, the extra force needed to compensate W/W*cos(climbangle) i.e., a slightly higher vertical force, comes for 90% from a slight increase in "lift", by means of a slightly higher AoA.
Only 10% will come from the vertical component of the increased thrust - which will have increased slightly to compensate for the increased drag from the slightly higher AoA (second order effect).
It will have changed by T*sin(climbangle), which will still be far less than the "lift" effect.
Yes, I know I'm sticking my neck out... especially since I promised to look at it this weekend but had other things to do...
Just to clarify, when I talk about a "typical" aircraft and a "typical" climb, I'm talking about a Cessna, or a Boeing, where lift/weight is about 10 times thrust/drag and a climb angle of something like 5° to 10° at the most.
Talking about vertical climbs confuses the issue, since by then you need engine power/thrust sufficient to maintain steady vertical flight, and the aerdynamics are no longer the same at all.
CJ