Question

1) (1+tg²alpha)cos²alpha 2) tg alpha + tg beta дробь ctg alpha + ctg beta 3) cos² alpha- ctg² alpha дробь sin² alpha - tg² alpha

Answer 1

$(1+tg^2alpha)cos^2alpha=cos^2alpha+sin^2alpha=1;\\ frac{tgalpha+tgbeta}{ctgalpha+ctgbeta}=frac{frac{sinalpha cosbeta + cosalpha sinbeta}{cosalpha cosbeta}}{frac{sinbeta cosalpha+cosbeta sinalpha}{sinalpha sin beta}}=frac{frac{sin(alpha + beta)}{cosalpha cosbeta}}{frac{sin(alpha+beta)}{sinalpha sinbeta}}=frac{sin(alpha +beta)}{cosalpha cosbeta}*frac{sinalpha sinbeta}{sin(alpha + beta)}=tgalpha tgbeta;$

$frac{cos^2alpha-ctg^2alpha}{sin^2alpha-tg^2alpha}=frac{frac{sin^2alpha cos^2alpha-cos^2alpha}{sin^2alpha}}{frac{sin^2alpha cos^2alpha-sin^2alpha}{cos^2alpha}}=frac{-cos^2alpha(1-sin^2alpha)}{sin^2a}*frac{cos^2alpha}{-sin^2alpha(1-cos^2alpha)}=\=frac{-cos^2alpha(cos^2alpha)}{sin^2alpha}*frac{cos^2alpha}{-sin^2alpha(sin^2alpha)}=frac{cos^6alpha}{sin^6alpha}=ctg^6alpha$