Math help needed - how do i prove : [cot^2(x)]/1+cscx = (1-sinx)/sinx

cot^2x /(1+cscx)
cos^2x/sin^2x /(1+1/sinx)
1-sin^2x/sin^2x /sinx/sinx + 1/sinx
(1-sinx)(1+sinx)/sin^2x/sinx+1/sinx
(1-sinx)(1+sinx)/sin^2x *(sinx/sinx+1)
(1-sinx)(1+sinx)(sinx)/sin^2x(sinx+1)
(1-sinx)/sinx ANSWER
hope this helps :)

There is more than one way to solve this:

You could:
Multiply the numerator and the denominator by 1 - csc x
You know that csc^2 (x) = cot^2(x)+ 1 = csc^2 (x) which means that 1 - csc^(x) = - cot^2(x)

Then
[cot^2(x)]/1+cscx

=[cot^2(x)] [1-cscx]/(1+cscx)(1-cscx)

= [cot^2(x)][1-cscx]/1-csc^2x

= [cot^2(x)][1-cscx]/1-csc^2x

= [cot^2(x)][1-cscx]/(-cot^2x)

= 1-cscx/(-1)

= cscx -1

= (1/sinx) - 1

= 1/sinx - sinx/sinx (to get common denominators)

= (1-sinx) / sinx

cot^2(x)/1+cosecx
= cosec^2(x)-1 / cosecx +1
= cosec x-1
=(1/sinx) -1
=(1-sinx)/sinx