Express cosA in terms of t when tan(A/2)=t
i checked the solution but i don't understand it.
solution:
using the half-angle formula for tan, tan^2(A/2)=(1-cosA)/(1+cosA)
therefore cosA=(1-tan^2(A/2))/(1+tan^2(A/2))=(1-t^2)/(1+t^2)
what i don't understand is how they get cosA in terms of tan^2(A/2).
how i would solve it is to multiply both sides of tan^2(A/2)=(1-cosA)/(1+cosA) with (1+cosA), but that won't get me cosA=(1-tan^2(A/2))/(1+tan^2(A/2))
sorry, i don't get it >.<
the question is this:
what i dont understand is this:
question 2.bmp