Question Number 176211 by adhigenz last updated on 15/Sep/22
Answered by mahdipoor last updated on 15/Sep/22
$$\mathrm{2021}\frac{{cos}\left({a}+{b}\right)}{{cosa}}{sinb}=\frac{{sina}}{{cosa}} \\ $$$$\mathrm{2021}\frac{{cosa}.{cosb}−{sina}.{sinb}}{{cosa}}{sinb}={tana} \\ $$$$\mathrm{2021}{cosb}.{sinb}−\mathrm{2021}{tana}.{sin}^{\mathrm{2}} {b}={tana} \\ $$$$\Rightarrow{tana}=\frac{\mathrm{2021}{cosb}.{sinb}}{\mathrm{1}+\mathrm{2021}{sin}^{\mathrm{2}} {b}}={f}\left({b}\right) \\ $$$${max}\:{f}\:=\:{max}\:{tana} \\ $$$$\begin{cases}{{cosb}.{sinb}=\frac{{sin}\mathrm{2}{b}}{\mathrm{2}}}\\{{sin}^{\mathrm{2}} {b}=\frac{\mathrm{1}−{cos}\mathrm{2}{b}}{\mathrm{2}}}\end{cases}\Rightarrow \\ $$$${f}\left({b}\right)=\frac{{sin}\left(\mathrm{2}{b}\right)}{\frac{\mathrm{2023}}{\mathrm{2021}}−{cos}\left(\mathrm{2}{b}\right)}\:\:\:\:\:\:\left({get}\frac{\mathrm{2023}}{\mathrm{2021}}={c}\right) \\ $$$$\frac{{df}}{{db}}=\mathrm{0}\:\Rightarrow\:{cos}\left(\mathrm{2}{b}\right)=\frac{\mathrm{1}}{{c}}\:\:\Rightarrow\:{sin}\left(\mathrm{2}{b}\right)=\frac{\sqrt{{c}^{\mathrm{2}} −\mathrm{1}}}{{c}} \\ $$$${max}\:{f}=\frac{\sqrt{{c}^{\mathrm{2}} −\mathrm{1}}/{c}}{{c}−\mathrm{1}/{c}}=\frac{\sqrt{{c}^{\mathrm{2}} −\mathrm{1}}}{{c}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{2021}\sqrt{\mathrm{2022}}}{\mathrm{4044}} \\ $$