Если tg(L+пи/4)=-1/3.Найдите значение ctg2L.Помогите пожалуйста.

[latex]tg(L+\frac{\pi}{4})=\frac{sin(2L+\frac{\pi}{2})}{cos(2L+\frac{\pi}{2})+1} = -\frac{1}{3}\\ 3sin(2L+\frac{\pi}{2})=-(cos(2L+\frac{\pi}{2})+1)\\ 3cos2L=-(-sin2L+1)\\ 3cos2L=sin2L-1\\ 3(2cos^2L-1)=2sinL*cosL-1\\ 6cos^2L-3=2\sqrt{1-cos^2L}*cosL-1\\ cosL=t\\ 6t^2-3=2\sqrt{1-t^2}*t-1\\ t=\frac{\sqrt{2}}{2}\\ L=\frac{\pi}{4}\\ \\ t=\frac{-\sqrt{5}}{5}\\ cosL=\frac{-\sqrt{5}}{5}\\ sinL=\frac{\sqrt{20}}{5}}\\ ctgL=-\frac{1}{2}\\ L=arctg(-0.5)\\ ctg2L=ctg(2arctg(-0.5))=-\frac{3}{4}[/latex]