Question

sina/sin^3a+3cos^3a если tg=2

Answer 1

$\displaystyle\bf\\tg\alpha =2\\\\tg\alpha =\frac{Sin\alpha }{Cos\alpha } \\\\\frac{Sin\alpha }{Cos\alpha } =2 \ \ \ \Rightarrow \ \ \boxed{Sin\alpha =2Cos\alpha} \\\\\\\frac{Sin\alpha }{Sin^{3}\alpha+3Cos^{3} \alpha } =\frac{2Cos\alpha }{(2Cos\alpha )^{3} +3Cos^{3} \alpha } =\frac{2Cos\alpha }{8Cos^{3}\alpha +3Cos^{3} \alpha } =\\\\\\=\frac{2Cos\alpha }{11Cos^{3}\alpha }=\frac{2}{11Cos^{2} \alpha }$