Question Number 148773 by mim24 last updated on 31/Jul/21
Answered by som(math1967) last updated on 31/Jul/21
$$\frac{\mathrm{1}}{\boldsymbol{{sin}}\mathrm{20}}\:+\frac{\sqrt{\mathrm{3}}}{\boldsymbol{{cos}}\mathrm{20}} \\ $$$$=\frac{\boldsymbol{{cos}}\mathrm{20}+\sqrt{\mathrm{3}}\boldsymbol{{sin}}\mathrm{20}}{\boldsymbol{{sin}}\mathrm{20}\boldsymbol{{cos}}\mathrm{20}} \\ $$$$=\frac{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{cos}}\mathrm{20}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\boldsymbol{{sin}}\mathrm{20}\right)}{\boldsymbol{{sin}}\mathrm{20}\boldsymbol{{cos}}\mathrm{20}} \\ $$$$=\frac{\mathrm{2}.\mathrm{2}\left(\boldsymbol{{cos}}\mathrm{60}\boldsymbol{{cos}}\mathrm{20}+\boldsymbol{{sin}}\mathrm{60}\boldsymbol{{sin}}\mathrm{20}\right)}{\mathrm{2}\boldsymbol{{sin}}\mathrm{20}\boldsymbol{{cos}}\mathrm{20}} \\ $$$$=\frac{\mathrm{4}\boldsymbol{{cos}}\left(\mathrm{60}−\mathrm{20}\right)}{\boldsymbol{{sin}}\mathrm{40}} \\ $$$$=\frac{\mathrm{4}\boldsymbol{{cos}}\mathrm{40}}{\boldsymbol{{sin}}\mathrm{40}}=\mathrm{4}\boldsymbol{{cot}}\mathrm{40}=\mathrm{4}\boldsymbol{{tan}}\left(\mathrm{90}−\mathrm{40}\right)=\mathrm{4}\boldsymbol{{tan}}\mathrm{50} \\ $$