More Trig Identities?

Question

Prove each identity.

(1) csc^2(x/2) = 2/(1 - cos(x))

(2) tan(x/2) = csc(x) - cot(x)

Answer 1

(1) RHS
= 2/(1-cos(x))
= 2/(1-cos^2(x/2) + sin^2(x/2))
= 2/(sin^2(x/2) + sin^2(x/2))
= 1/sin^2(x/2)
= csc^2(x/2)
= LHS

(2) RHS
= csc(x) - cot(x)
= 1/sin(x) - cos(x)/sin(x)
= (1-cos(x))/sin(x)
= (1-(1-2sin^2(x/2))) / (2sin(x/2)cos(x/2))
= 2sin^2(x/2) / (2sin(x/2)cos(x/2))
= sin(x/2)/cos(x/2)
= tan(x/2)
= LHS

Answer 2

as cos x = 1 - 2 sin ^2 x/2 , then 2 sin^2 x/2 =1-cosx

then sin^2 x/2 = (1 - cos x ) / 2

as 1/sin x = csc x , then
csc^2(x/2) = 2/(1 - cos(x)) q.e.d

2.lhs 1/sinx - cos /sin x = 1-cos x / sin x x/2)/2
= 1 - (1 - 2 sin^2 x/2)/ 2 sin x/2 cos x/2 = sin x/2 / cos x/2
= tan x/2 qed

Answer 3

(1) csc^2(x/2) = 2/(1 - cos(x))
1/sin^2 (x/2)=2/{(sin^2 (x/2)+cos^2(x/2)-cos^2 (x/2)+sin^2(x/2)}

1/sin^2 (x/2)=2/{2 (sin^2 (x/2)}

1/sin^2 (x/2)=1/ (sin^2 (x/2)

LHS=RHS

Remember 1=?
cosx=?
cos2x=?