Odpowiedź i szczegółowe wyjaśnienie:
[tex]\dfrac{2sinxcosx-1}{sin^2x-cos^2x}=\dfrac{1-tgx}{1+tgx}[/tex]
Założenia:
[tex]sin^2x-cos^2x\neq0 =>\ sin^2x\neq cos^2x\\\\1+tgx\neq0=>\ tgx\neq-1[/tex]
[tex]\dfrac{2sinxcosx-1}{sin^2x-cos^2x}=\dfrac{1-tgx}{1+tgx}\\\\\\\\L=\dfrac{2sinxcosx-1}{sin^2x-cos^2x}\\\\\\L=\dfrac{2sinxcosx-(sin^2x+cos^2x)}{(sinx-cosx)(sinx+cosx)}}\\\\\\L=\dfrac{2sinxcosx-sin^2x-cos^2x)}{(sinx-cosx)(sinx+cosx)}}\\\\\\L=\dfrac{-sinx +2sinxcosx-cos^2x}{(sinx-cosx)(sinx+cosx)}}\\\\\\L=\dfrac{-(sin^2x-2sinxcosx+cos^2x)}{(sinx-cosx)(sinx+cosx)}}\\\\\\L=\dfrac{-(sinx-cosx)^2}{(sinx-cosx)(sinx+cosx)}}\\\\\\L=\dfrac{-(sinx-cosx)}{(sinx+cosx)}}\\\\\\[/tex]
[tex]L=\dfrac{cosx-sinx}{sinx+cosx}}\\\\\\L=\dfrac{\frac{cosx}{cosx}-\frac{sinx}{cosx}}{\frac{sinx}{cosx}+\frac{cosx}{cosx}}\\\\\\L=\dfrac{1-tgx}{tgx+1}\\\\\\L=\dfrac{1-tgx}{1+tgx}\\\\\\L=P[/tex]
