Question

Решите пожалуйста )))))))))))))))

Answer 1

$1.;frac{sin^2alpha-1}{1-cos^2alpha}=frac{sin^2alpha-sin^2alpha-cos^2alpha}{sin^2alpha+cos^2alpha-cos^2alpha}=frac{-cos^2alpha}{sin^2alpha}=-tg^2alpha\alpha=fracpi4Rightarrow -tg^2alpha=-(1)^2=-1\\2.;cos^2alpha+ctg^2alpha+sin^2alpha=1+ctg^2alpha=1+frac{cos^2alpha}{sin^2alpha}=frac{sin^2alpha+cos^2alpha}{sin^2alpha}=\=frac1{sin^2alpha}\alpha=fracpi6Rightarrowfrac1{sin^2alpha}=frac1{left(frac12right)^2}=frac1{frac14}=4$

$3.;frac1{cos^2alpha}-1=frac{1-cos^2alpha}{cos^2alpha}=frac{sin^2alpha+cos^2alpha-cos^2alpha}{cos^2alpha}=frac{sin^2alpha}{cos^2alpha}=tg^2alpha\alpha=fracpi3Rightarrow tg^2alpha=(sqrt3)^2=3\\4.;cos^2alpha+tg^2alpha ctg^2alpha+sin^2alpha=1+frac{sin^2alpha}{cos^2alpha}cdotfrac{cos^2alpha}{sin^2alpha}=1+1=2$
Последнее равенство справедливо для любых α