Question 217331
Prove this identity


{{{ (sin2x + cos2x -1) / (sin2x + cos2x +1) }}} = {{{ (1 - tanx )/ (1+cotx) }}}


Step 1.  {{{sin2x=2cosxsinx}}}, {{{cos2x=(cosx)^2-(sinx)^2=1-2(sinx)^2}}}, {{{(sin x)^2+(cos x)^2=1}}}, and {{{tanx=sinx/cosx}}}


Step 2.  Let's work the numerator:


{{{ sin2x + cos2x -1 =  2cosxsinx+1-2(sinx)^2-1}}}


{{{ sin2x + cos2x-1 = 2 cosxsinx-2(sin x)^2}}}


Factor{{{ 2(sin x)^2}}}


{{{ sin2x + cos2x-1 = (cosx/sinx- 1)2(sin x)^2}}}


{{{ sin2x+cos2x-1=(cotx-1)2(sinx)*2}}}


Step 3.  Let's work with the denominator:  {{{sin2x + cos2x +1}}}


{{{ sin2x + cos2x + 1 =  2cosxsinx+2(cosx)^2-1+1}}}


{{{ sin2x + cos2x+1 = 2 cosxsinx+2(cos x)^2}}}


Factor{{{ 2(cos x)^2}}}


{{{ sin2x + cos2x+1 = (sinx/cosx+1)2(cos x)^2}}}


{{{ sin2x+cos2x+1=(tanx+1)2(cosx)^2}}}



Step 4.  Now combine the equations Step 3 and 4


{{{ (sin2x + cos2x -1) / (sin2x + cos2x +1) }}} = {{{((cotx-1)2(sinx)*2)/(tanx+1)2(cosx)^2)}}}


But {{{(sinx)^2/(cox)^2=tanx/cotx}}}


{{{ (sin2x + cos2x -1) / (sin2x + cos2x +1) }}} = {{{(cotx-1)*tanx/(tanx+1)cotx}}}


{{{ (sin2x + cos2x -1) / (sin2x + cos2x +1) }}} = {{{(1-tanx)/(1+cotx)}}}


Step 5  ANSWER:  {{{ (sin2x + cos2x -1) / (sin2x + cos2x +1) }}} = {{{(1-tanx)/(1+cotx)}}}