Question

Выражение ( sin4a+cos4a·ctg2a )/(sin4a-cos4a·tg2a )

Answer 1

$\frac{\sin 4 \alpha+\cos 4\alpha\cdot ctg2\alpha }{\sin4\alpha-\cos 4\alpha\cdot tg2\alpha} = \frac{2\sin2\alpha\cos2\alpha+(1-2\sin^22\alpha)\cdot \frac{\cos 2\alpha}{\sin2\alpha} }{2\sin2\alpha\cos2\alpha-(2\cos^22\alpha-1)\cdot \frac{\sin2\alpha}{\cos2\alpha} } = \\ \\ = \frac{2\sin2\alpha\cos2\alpha+ \frac{\cos2\alpha}{\sin2\alpha}-2\sin2\alpha\cos2\alpha }{2\sin2\alpha\cos2\alpha-2\cos2\alpha\sin2\alpha+ \frac{\sin2\alpha}{\cos2\alpha} } =\dfrac{ \frac{\cos2\alpha}{\sin2\alpha} }{ \frac{\sin2\alpha}{\cos2\alpha} }= \frac{\cos^22\alpha}{\sin^22\alpha} =ctg^22\alpha$