Verify this trig identity csc x cot x2 1cosx 1 cos xSolu

Verify this trig identity. (csc x - cot x)^2 = (1-cosx / 1 + cos x)

given that

(csc x - cot x)^2 = (1- cosx / 1+cosx )

L.H.S = (csc x - cost x)^2

but csc x = 1/sin x and cot x = cos x /sin x

L.H.S = (1/sin x - cosx /sinx)^2

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

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

= (1- cos x)^2 / 1- cos^2 x [ by using sin^2 x + cos^2 x=1 ]

= (1- cos x)^2 / (1+ cos x) (1-cosx)

=[ (1-cosx) (1-cosx) ] / [ (1+ cos x) (1-cosx)]

= 1-cos x / 1+ cosx

L.H.S = R.H.S

hence proved