Question

Выражение: ctg^2x-cos^2x-ctg^2xcos^2x решить . cрочно!

Answer 1

$\cot^2x-\cos^2x-\cot^2x\cdot\cos^2x=\dfrac{\cos^2x}{\sin^2x}-\cos^2x-\dfrac{\cos^4x}{\sin^2x}=\\\\=\dfrac{\cos^2x-\cos^2x\cdot\sin^2x-\cos^4x}{\sin^2x}=\dfrac{\cos^2x\cdot (1-\sin^2x-\cos^2x)}{\sin^2x}=\\\\=\dfrac{\cos^2x\cdot (\cos^2x-\cos^2x)}{\sin^2x}=\dfrac{\cos^2x\cdot 0}{\sin^2x}=0$

Answer 2

$\frac{cos^2x}{sin^2x} - \frac{cos^2x*sin^2x}{sin^2x}- \frac{cos^4x}{sin^2x}= \\ = \frac{cos^2x-cos^2x+cos^4x-cos^4x}{sin^2x}= \frac{0}{sin^2x}=0$