Question

Să se arate că pe domeniul de definiție au loc egalitățile:
$\frac{ {sin}^{2}x }{sinx - cosx} - \frac{sinx + cosx}{ {tg}^{2}x - 1 } = sinx + cosx$

Answer 1

   
[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]