[tex]E = \dfrac{m+\sin 2x+m\cos 2x}{1+m\sin 2x-\cos 2x} = \dfrac{m(1+\cos 2x)+\sin 2x}{m\sin 2x+1-\cos 2x}= \\ \\ =\dfrac{m(1+\cos^2 x-\sin ^2 x)+2\sin x\cos x}{2m\sin x \cos x+1-\cos^2 x+\sin ^2 x}= \\ \\ =\dfrac{m(1+\cos^2 x-1+\cos^2 x+2\sin x\cos x)}{2m\sin x\cos x-1+\sin^2 x+\sin^2 x} = \\ \\ = \dfrac{2m\cos^2 x+2\sin x\cos x}{2m\sin x\cos x+2\sin^2 x} = \dfrac{2\cos x(m \cos x +\sin x)}{2\sin x(m\cos x+\sin x)} = \dfrac{\cos x}{\sin x} = \\ \\ = \text{ctg}\,x,\quad \forall m[/tex]
