433(1,3) 434(1,3) спасибо большое!

Решите задачу:

$1)\; \; tga=2\\\\\frac{3sina-5cosa}{4sina+cosa} =[\, \frac{:cosa}{:cosa}\, ]= \frac{3tga-5}{4tga+1} = \frac{6-5}{8+1} =\frac{1}{9} \\\\2)\; \; tga=2\\\\ \frac{sin^3a-2cos^3a}{2sin^3a+cos^3a} = [\; \frac{:cos^3a}{:cos^3a} \; ]= \frac{tg^3a-2}{2tg^3a+1} = \frac{8-2}{16+1}=\frac{6}{17}\\\\3)\; \; ctga=-2\\\\\frac{2sina+3cosa}{5sina-cosa} =[\; \frac{:sina}{:sina} \; ]= \frac{2+3ctga}{5-ctga}= \frac{2-6}{5+2}= \frac{-4}{7}$

$4)\; \; ctga=-2\\\\ \frac{cosa+2sina}{sin^3a-2cos^3a} =[\; \frac{:sin^3a}{:sin^3a} \; ]= \frac{ctga\cdot \frac{1}{sin^2a}+2\cdot \frac{1}{sin^2a}}{1-2ctg^3a} =\\\\= \frac{ctga\cdot (1+ctg^2a)+2\cdot (1+ctg^2a)}{1-2ctg^3a} = \frac{(1+ctg^2a)\cdot (2+ctga)}{1-2ctg^3a} =\\\\= \frac{(1+4)\cdot (2-2)}{1-16} = 0$