HWD
You are calculating the lift reduction from the lift, it is actually a component of the drag (as it is the upward component of the drag, which is upslope, that reduces the lift requirement). Shame on you for taking lift in kg
(my physics tutors would spin in their graves at the units JAA force us to use, if they were dead), but to simplify I will keep it the same. The drag would be about 1000 / 11 = 91 kg. The component of this upward would be 91 x sin (5 degrees). This is about 8 kg. To complicate the issue the forward slope of the lift vector means that roughly 4 kg extra lift is required, meaning a net reduction of the order 0.4% in the lift required.
Again I have to add "I think". This is well beyond the ATPL syllabus, drawing on half-remembered first-year mechanics. I may well have missed out another factor, as at these levels approximations start to show their flaws.
MTV
Don't worry - not only have I done the whole course but I taught it too, although that was some time ago. This is beyond what you will need to know!
I find it interesting though. Sad
