[tex]\displaystyle\\
\text{Trebuie sa aratam ca are loc egalitatea:}\\\\
\frac{\sin^2x}{\sin x-\cos x}-\frac{\sin x + \cos x}{\text{tg}^2x-1}=\sin x+\cos x\\\\
\text{Rezolvare:}\\\\
\frac{\sin^2x}{\sin x-\cos x}-\frac{\sin x + \cos x}{\text{tg}^2x-1}=\\\\
=\frac{\sin^2x}{\sin x-\cos x}-\frac{\sin x + \cos x}{ \dfrac{\sin^2x}{\cos^2x} -1}=\\\\\\
=\frac{\sin^2x}{\sin x-\cos x}-\frac{\sin x + \cos x}{ \dfrac{\sin^2x-\cos^2x}{\cos^2x}}=
[/tex]
[tex]\displaystyle\\
=\frac{\sin^2x}{\sin x-\cos x}-\frac{\cos^2x(\sin x + \cos x)}{ \sin^2x-\cos^2x}=\\\\
=\frac{\sin^2x}{\sin x-\cos x}-\frac{\cos^2x(\sin x + \cos x)}{(\sin x + \cos x)(\sin x-\cos x)}=\\\\
=\frac{\sin^2x}{\sin x-\cos x}-\frac{\cos^2x}{\sin x-\cos x}=\\\\
=\frac{\sin^2x-\cos^2x}{\sin x-\cos x}=\\\\
=\frac{ (\sin x + \cos x)(\sin x-\cos x)}{\sin x-\cos x}= \boxed{\sin x + \cos x}\\\\
\text{cctd}[/tex]
