multiply by 2sin[A] on both sides of cos[A]/a=sin[A]/b
you will have smthng roughly like this:
2sin[A] x cos[A]/a=2sin[A] x sin[A]/b
since,2sin[A].cos[A] is the formula of sin2[A],lets put sin2[A]
=>sin2[A]/a=2sin^2[A]/b
cross multiply:
bsin2[A]=a2sin^2[A]
from the formula cos2B=1-2sin^2B we can get 2sin^2B=1-cos2B lets put this on the RHS
bsin2[A]=a.(1-cos2[A])
open the bracket:
bsin2[A]=a-a.cos2[A]
=>bsin2[A]+acos2[A]=a
you can prove the same for 'b' by multiplying and dividing by 2cos[A]
hope it is helpful
