Question Number 86948 by gny last updated on 01/Apr/20
$$\frac{−{sin}\alpha+\mathrm{3}{cos}\alpha}{\mathrm{4}{cos}\alpha+\mathrm{3}{sin}\alpha} \\ $$$$ \\ $$$${tg}\alpha=−\mathrm{3} \\ $$$${pls}\:{help}\:{me}\:{sir} \\ $$
Commented by Prithwish Sen 1 last updated on 01/Apr/20
$$\mathrm{divide}\:\mathrm{n}_{\mathrm{r}} \:\mathrm{and}\:\mathrm{d}_{\mathrm{r}} \:\mathrm{by}\:\mathrm{cos}\alpha\:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{−\mathrm{tan}\alpha+\mathrm{3}}{\mathrm{4}+\mathrm{3tan}\alpha}\:=\:\frac{\mathrm{3}+\mathrm{3}}{\mathrm{4}−\mathrm{9}}\:=\:−\frac{\mathrm{6}}{\mathrm{5}} \\ $$