Срочно!! Докажите, что

Решите задачу:

$\displaystyle \tan^2(x) - \sin^2(x)= \tan^2(x) \cdot \sin^2(x) \\\\ \frac{\sin^2x}{\cos^2 x}-\sin^2x= \frac{\sin^2x}{\cos^2 x}\cdot \sin^2 x\\\\\ \frac{\sin^2x-\cos^2x\cdot\sin^2 x}{\cos^2 x}= \frac{\sin^4x}{\cos^2 x} \\\\ \sin^2x-\cos^2x\cdot\sin^2 x=\sin^4x\\\\\sin^2x(1-\cos^2 x)=\sin^4x\\\\\sin^2x\cdot \sin^2 x=\sin^4x\\\\\sin^4x=\sin^4 x\\\\ 0=0$