Question Number 187375 by Rupesh123 last updated on 16/Feb/23
Commented by mokys last updated on 16/Feb/23
$$\frac{\frac{{cosx}−{sinx}}{{cosx}}}{\frac{{sinx}−{cosx}}{{sinx}}}\:\:=\:\frac{−{secx}}{{cscx}}\:=\:−{tanx}\:\neq\:\mathrm{2}{sinx} \\ $$
Commented by Frix last updated on 17/Feb/23
$$\mathrm{Wrong}. \\ $$$$−\mathrm{tan}\:{x}\:=\mathrm{2sin}\:{x} \\ $$$$\mathrm{0}\leqslant{x}<\mathrm{2}\pi \\ $$$${x}\in\left\{\mathrm{0},\:\frac{\mathrm{2}\pi}{\mathrm{3}},\:\pi,\:\frac{\mathrm{4}\pi}{\mathrm{3}}\right\} \\ $$