Question Number 160373 by HongKing last updated on 28/Nov/21
$$\mathrm{Compare}\:\mathrm{it}: \\ $$$$\frac{\mathrm{1}\:-\:\mathrm{sin}\:\left(\mathrm{10}°\right)}{\mathrm{cos}\:\left(\mathrm{10}°\right)}\:\:\:\:\:\mathrm{and}\:\:\:\:\:\mathrm{1} \\ $$
Answered by ghimisi last updated on 28/Nov/21
$$\frac{{sin}\mathrm{90}−{sin}\mathrm{10}}{{cos}\mathrm{10}}=\frac{\mathrm{2}{sin}\mathrm{40}{cos}\mathrm{50}}{{sin}\mathrm{80}}= \\ $$$$\frac{\mathrm{2}{sin}\mathrm{40}{sin}\mathrm{40}}{\mathrm{2}{sin}\mathrm{40}{cos}\mathrm{40}}={tg}\mathrm{40}<{tg}\mathrm{45}=\mathrm{1} \\ $$
Commented by HongKing last updated on 28/Nov/21
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much},\:\mathrm{cool},\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser} \\ $$
Answered by mr W last updated on 29/Nov/21
$${f}\left({x}\right)=\frac{\mathrm{1}−\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${f}'\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}\:{x}}\:<\:\mathrm{0}\:{for}\:{x}<\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$${that}\:{means}\:{f}\left({x}\right)\:{is}\:{strictly}\:{decreasing} \\ $$$${for}\:\mathrm{0}\leqslant{x}\leqslant\frac{\mathrm{3}\pi}{\mathrm{2}}.\:{or} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}−\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}<{f}\left(\mathrm{0}\right)=\mathrm{1}\:{for}\:{any}\:{x} \\ $$$$\mathrm{0}<{x}\leqslant\frac{\mathrm{3}\pi}{\mathrm{2}}\:{including}\:{x}=\mathrm{10}° \\ $$
Commented by HongKing last updated on 29/Nov/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{cool} \\ $$