Вычислите ........................................

Решите задачу:

$1)\\tg\alpha+ctg\alpha={1\over sin\alpha cos\alpha}=3\\sin\alpha cos\alpha={1\over3}\\sin^2\alpha =x, cos^2\alpha=y\\\\xy={1\over9}\\x+y=1\\\\x^2+y^2=1-{2\over9}={ 7\over9}\\\\tg^2\alpha+ctg^2\alpha={sin^4\alpha+cos^4\alpha\over sin^2\alpha cos^2\alpha}=9(sin^4\alpha+cos^4\alpha)=7\\\\2)\\tg\alpha=-3\Rightarrow sin\alpha=-3cos \alpha \\\\\frac{3\sin\alpha-4cos \alpha }{ 5sin \alpha +6cos \alpha } ={-9-4\over-15+6}={13\over9}$