Question

Решить)) 1) tg^2φ(sin^2φ-1) 2) sin γ (tg γ + ctg γ) - cos γ/sin γ

Answer 1

Для удобства я вместо букв фи и гамма буду везде использовать x

$1)\displaystyle \,\,tg^2x(sin^2x-1)\quad \quad \quad \bigg(tgx=\frac{sinx}{cosx}\bigg)\\\\-\frac{sin^2x}{cos^2x}(1-sin^2x)\\\\-\frac{sin^2x}{cos^2x}*cos^2x\\\\\boxed{-sin^2x}\\\\\\\\2)\frac{sinx(tgx+ctgx)-cosx}{sinx} \quad \quad \quad \bigg(ctgx=\frac{cosx}{sinx}\bigg)\\\\\frac{sinx\bigg(\frac{sinx}{cosx}+\frac{cosx}{sinx}\bigg)-cosx}{sinx}\\\\\frac{\frac{sin^2x}{cosx}+cosx-cosx}{sinx}\\\\\frac{\frac{sin^2x+cos^2x}{cosx}-cosx}{sinx}\\\\\frac{\frac{1}{cosx}-cosx}{sinx}\\\\\frac{1-cos^2x}{sinx*cosx}$

$\displaystyle \frac{sin^2x}{sinx*cosx}\\\\\frac{sinx}{cosx}\\\\\boxed{tgx}$