OK, hands up to whoever said the real issue is the lift vector being turned back over (climbangle) degrees.
Everything else is secondary, except that the thrust angle is turned up over the same angle.
If you feel like doing some scribbling, try with an aircraft of 2000 kg, a CL/CD of 10, and a climb angle of about 6° (sine = 0.1, cosine = 0.995, so practically =1).
I'll try and clean up my scribbles and post them, but here goes in the meantime.
In level flight, drag and thrust are 200 kgf.
Start a steady climb at abt 6° climb angle (nothing excessive, that's 2000 ft/min at 200 kts).
Now you have to drag 2000 kgf upstairs at W*sin(ca), i.e., 200 kgf more.
Aye, there's the rub.....
So T now becomes 400 kgf, and the vertical component = T*sin(ca) = 40 kgf.
Since we're talking steady state, the component from the lift will have to be reduced by the same 40 kgf. But.... L=W*cos(ca) is only about 4kgf less than W. So we have to reduce the lift by another 36kgf to get back to a steady state.
Since 2000 kg, CL/CD=10, and a climb at 6° are hardly excessive, it would seem Rainboe's intuition about looking at a steep climb was right after all....
Now to question 2.....