if cos = L, then
sin = √(1-L^2)
csc=1/√(1-L^2)
sec = 1/L
tan = √(1-L^2)/L
cot = L/√(1-L^2)
so you have
L/√(1-L^2) - 1/√(1-L^2)
--------------------------
1/L - √(1-L^2)/L
(L-1)/√(1-L^2)
-------------------
(1 - √(1-L^2))/L
L(L-1)
---------------------
√(1-L^2)(1-√(1-L^2))
Not sure how far you want to play with it.
Leaving stuff as trig function, you get something as intractable:
(cos-1)/(sin+1) * cot
Given that
cosZ=L
where z is an acute angle.find and expression of (cotz-cosecz)/(secz+tanz)?
plz help help help!!!! Show work
1 answer