Question

3cosa-4sina/4sina+cosa, если tga=1

Answer 1

$tga=1\\\\ \frac{3cosa-4sina}{4sina+cosa} =\frac{cosa\cdot (3-4tga)}{cosa\cdot (4tga+1))}=\frac{3-4tga}{4tga+1}=\frac{3-4}{4+1}=\frac{-1}{5}$