trig identities?

Question

how can i prove this?

sin x/1-cosx + 1-cosx/sinx = 2cscx

Answer 1

sinx/(1-cosx) + (1-cosx)/sinx

sinx/(1-cosx) * sinx/sinx + (1-cosx)/sinx * (1-cosx)/(1-cosx)

sin^2x/[sinx(1-cosx)] + [(1-cosx)^2]/[sinx(1-cosx)]

[sin^2x + (1-cosx)^2]/[sinx(1-cosx)]

[sin^2x + 1 - 2cosx + cos^2x]/[sinx(1-cosx)]

[1 + 1 - 2cosx]/[sinx(1-cosx)] <-- sin^2x + cos^2x = 1

[2 - 2cosx]/[sinx(1-cosx)]

2[1 - cosx]/[sinx(1-cosx)]

2/sinx

2cscx <----cscx = 1/sinx

Answer 2

Brackets have not been used so have to guess that equation is not as given but should read as:-

sin x/(1 - cos x) + (1 - cos x) / sin x = 2 cosec x

LHS
[ sin ² x + (1 - cos x)² ] / [( sin x )(1 - cos x ) ]
[sin ² x + 1 - 2 cos x + cos ² x ]/(sin x)(1 - cos x)
[ 2(1 - cos x) ] / [ (sin x)(1 - cos x) ]
2 / sin x
2 cosec x

RHS
2 cosec x

LHS = RHS

Answer 3

sin x /(1 - cos x) + (1 - cos x )/ sin x = 2 csc x
sin x /(1 - cos x) (1 + cos x) / (1 + cos x) + (1 - cos x )/ sin x =
sin x (1 + cos x) /(1 - cos ² x) + (1 - cos x )/ sin x =
sin x (1 + cos x) /sin ² x + (1 - cos x )/ sin x =
(1 + cos x) / sin x + (1 - cos x )/ sin x =
( 1 + cos x + 1 - cos x) / sin x =
2 / sin x =
2 csc x = same as the right