Question

Приведите к более простому виду

Answer 1

$\dfrac{Ctg^{2}\alpha-Cos^{2}\alpha}{tg^{2} \alpha-Sin^{2}\alpha}=\dfrac{\dfrac{Cos^{2} \alpha}{Sin^{2}\alpha}-Cos^{2}\alpha}{\dfrac{Sin^{2}\alpha}{Cos^{2}\alpha}-Sin^{2}\alpha} =\dfrac{\frac{Cos^{2}\alpha-Cos^{2}\alpha\cdot Sin^{2} \alpha}{Sin^{2}\alpha}}{\dfrac{Sin^{2} \alpha-Sin^{2}\alpha\cdot Cos^{2}\alpha }{Cos^{2}\alpha}}=\\\\=\dfrac{Cos^{2}\alpha\cdot(1-Sin^{2} \alpha)\cdot Cos^{2}\alpha}{Sin^{2}\alpha\cdot(1-Cos^{2}\alpha)\cdot Sin^{2}\alpha}=$

$=\dfrac{Cos^{4}\alpha\cdot Cos^{2}\alpha}{Sin^{4}\alpha\cdot Sin^{2}\alpha} =\dfrac{Cos^{6}\alpha}{Sin^{6}\alpha}=\boxed{Ctg^{6}\alpha}$