How to prove this identity

(1 + tanx)/(1+cotx) = (1 - tanx)/(cotx-1)

(1 + tanx)/(1+cotx)
= (1 + tanx) / (1 + 1/tanx)
= tanx (1 + tanx) / (tanx + 1)
= tanx
= 1/ cotx
= (1 -- tanx) / cotx(1 -- tanx)
= (1 - tanx) / (cotx -- 1)

Simplifying the LHS:

(cos(x) + sin(x))/cos(x) * sin(x)/(sin(x) + cos(x)) = tan(x)

Simplifying the RHS:

(1 - tan(x))/(cot(x) - 1) = (tan(x) - 1)/(1 - cot(x))

= (sin(x) - cos(x))/cos(x) * sin(x)/(sin(x) - cos(x))

= tan(x)

Hence both sides are equal.

(1+tanx) / (1+cotx) = (1+sinx/cosx) / (1+cosx/sinx) = ((cosx+sinx) / cosx) / ((sinx+cosx) / sinx) = sinx / cosx = tanx

(1 - tanx) / (cotx-1) = (1-sinx/cosx) / (cosx/sinx-1) = ((cosx-sinx) / cosx) / ((cosx-sinx) / sinx) = sinx / cosx = tanx

Therefore (1 + tanx)/(1+cotx) = (1 - tanx)/(cotx-1)