Utilizam relatiile sin2x=2sinxcosx si cos2x=cos^{2}x-sin^{2}x .
Inlocuim in prima relatie:
\frac{1-cos2x}{sin2x}=\frac{1-cos^{2}x+sin^{2}x}{2sinxcosx}=\frac{sin^{2}x+cos^{2}x-cos^{2}x+sin^{2}x}{2sinxcosx} =\frac{2sin^{2}x}{2sinxcosx} =\frac{sinx}{cosx} =tgx
Asemanator rezolvam a doua relatie:
\frac{sin2x}{1+cos2x} =\frac{2sinxcosx}{1+cos^{2}x-sin^{2}x}=\frac{2sinxcosx}{sin^{2}x+cos^{2}x+cos^{2}x-sin^{2}x}=\frac{2sinxcosx}{2cos^{2}x}= \frac{sinx}{cosx}=tgx
