Oh god, I can't even do the first question!! Help me:
Use the derivatives of cosecx and cotx to prove that
d[ln(cosecx + cotx)] /dx = -cosecx
Thank you if you can help.
let y=ln(cosecx + cotx) let u=cosecx+cotx y=ln u => dy/du=1/u du/dx= -cosecxcotx-cosec^2x dy/dx=dy/du.du/dx= (-coseccotx-cosec^2x)/(cosecx+cotx) =-cosecx(cotx+cosecx)/(cosecx+cotx)=-cosecx
let y=ln(cosecx + cotx) let u=cosecx+cotx y=ln u => dy/du=1/u du/dx= -cosecxcotx-cosec^2x dy/dx=dy/du.du/dx= (-coseccotx-cosec^2x)/(cosecx+cotx) =-cosecx(cotx+cosecx)/(cosecx+cotx)=-cosecx
Lol, I was this close |<--->| to doing it, I couldn't find the work so did it again, but you beat me to it.
Hehe. I did this paper last year. Got so nervous in the exam hall, cos it was the first question (apparently loadsa people struggled with it). I came back to it at the end though, and found it easy (since I was relaxed etc).