1) (1+tg²alpha)cos²alpha 2) tg alpha + tg beta дробь ctg alpha + ctg beta 3) cos² alpha-...

Решите задачу:

$(1+tg^2\alpha)cos^2\alpha=cos^2\alpha+sin^2\alpha=1;\\\\ \frac{tg\alpha+tg\beta}{ctg\alpha+ctg\beta}=\frac{\frac{sin\alpha cos\beta + cos\alpha sin\beta}{cos\alpha cos\beta}}{\frac{sin\beta cos\alpha+cos\beta sin\alpha}{sin\alpha sin \beta}}=\frac{\frac{sin(\alpha + \beta)}{cos\alpha cos\beta}}{\frac{sin(\alpha+\beta)}{sin\alpha sin\beta}}=\frac{sin(\alpha +\beta)}{cos\alpha cos\beta}*\frac{sin\alpha sin\beta}{sin(\alpha + \beta)}=tg\alpha tg\beta;$

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