how does ( 1- cos x / sin x) = ( sin x / 1 + cos x )?

sin²(x)+cos²(x) = 1
1-cos²(x) = sin²(x)

LHS
= (1-cos(x)) / sin(x)
= (1-cos(x))(1+cos(x)) / (sin(x)(1+cos(x))
= (1-cos²(x)) / (sin(x)(1+cos(x))
= sin²(x) / (sin(x)(1+cos(x))
= sin(x) / (1+cos(x))
= RHS

Before we start , the question is INCORRECT.

Should be shown as :-

(1 - cos x ) / sin x = sin x / (1 + cos x )
And NOT as given.

(1 - cos x) (1 + cos x) = sin ² x

1 - cos ² x = sin ² x

sin ² x = sin ² x as required

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

Hence, this is an identity.

What does sin x / 1 mean? I guess you made some error writing question, what is the correct one?

(1- cos x) / sinx

Multiplying both denominator and numerator by 1 + cos x

(1 - cos x)(1 + cos x) / sin x(1 + cos x)

= (1 - cos² x)/ sinx (1 + cos x) [because a² - b² = (a - b)(a + b)]
= sin² x / sin x(1 + cos x) [because 1 - cos²x = sin²x]
= sin x / (1 + cos x)

I'm going to start from the right side.

u have to multiply ( sin x / 1 + cos x ) by its conjugate, which is (1-cosx)/(1-cosx)
--> (sinx-sinxcosx) / 1 - cos squared x
--> sinx(1-cosx) / 1 - cos squared x
use formula cos squared x= 1-sin squared x
-->sinsquared x=1-cos squared x
sinx (1-cosx) / sin squared x
sinx cancel
and you're left with the left side
(1-cosx / sinx)