[TEX]\frac{(1-cosx)(1+cosx)}{(1-sinx)(1+sinx)} = \frac{(1-cosx)(1 +cos^2x +cosx)}{(1-sinx)(1+sin^2x +sinx} \\ cosx = 1 \\ \frac{1+cosx}{1+sinx} = \frac{1 +cos^2x +cosx}{1+sin^2x +sinx} \\ sin^2 x-cos^2 x + sin^2x.cosx - cos^2x.sinx = 0 \\ sinx = cosx \\ sin x +cosx + sinx.cosx = 0 \\ u = sinx +cosx \Rightarrow 2.sinx.cosx = u^2 -1 \\ 2u + u^2 -1 = 0[/TEX]