AMATHS IDENTITY PROVE (sin2x+cos2x-1)/(sin2x+cos2x+1) = (1-tanx)/(1+cotx)

Prove that (sin2x+cos2x-1)/(sin2x+cos2x+1) = (1-tanx)/(1+cotx)

Remember tanx =sinx/cosx and reverse for cotx

LHS
EXPAND TOP OUT
2sinx cosx+cos^2x-Sin^2-1 remember cos^2x+sin^2x=1
2sinxcosx+cos^2x-sin^2x-(cos^2x+sin^2x…
2sinxcosx - 2sin^2x
EXPAND BOTTOM OUT
2sinxcosx+cos^2x-Sin^2x +(cos^2x+sin^2x)
2sinxcosx +2cos^2x

divide top and bottom by 2sinxcosx
TOP 1-sinx/cosx
BOTTOM 1+cosx/sinx

so 1-tanx/1+cotx

(sin2x+cos2x-1)/(sin2x+cos2x+1) = (1-tanx)/(1+cotx)
LHS
=(sin2x+cos2x-1)/(sin2x+cos2x+1)
=(2sin x cos x + (cos^2 x-sin^2 x)-1)/(2sin x cos x+(cos^2 x-sin^2 x)+1)
=(2sin x cos x + cos^2 x-sin^2 x-1)/(2sin x cos x+cos^2 x-sin^2 x+1)
=(2sin x cos x +1- sin^2 x -sin^2 x-1)/(2sin x cos x+cos^2 x-1+ cos^2 x+1)
=(2sin x cos x - sin^2 x -sin^2 x)/(2sin x cos x+cos^2 x+ cos^2 x)
=(2sin x cos x - 2sin^2 x )/(2sin x cos x+2cos^2 x)
=2sin x (cos x - sin x )/ 2cos x ( sin x+cos x)
=2sin x cos x (1 - tan x )/ 2cos x sin x ( 1+cot x)
= (1 - tan x )/ ( 1+cot x)
=RHS

(2sinxcosx-(1-cos2x))/(2sinxcosx+2cos^2x…
(2sinxcosx-2sin^2x)/(2sinxcosx+2cos^2x…
2sinx(cosx-sinx)/2cosx(sinx+cosx)
sinx(1-tanx)/(sinx+cosx)
(1-tanx)/(sinx+cosx)/sinx
(1-tanx)/(1+tanx)