Question Number 207157 by hardmath last updated on 07/May/24
$$\mathrm{Find}:\:\:\:\frac{\mathrm{5}\:\mathrm{sin}\:\frac{\pi}{\mathrm{6}}}{\frac{\mathrm{1}}{\mathrm{tg}\:\mathrm{75}°}\:\:−\:\:\mathrm{tg}\:\mathrm{75}°}\:\:=\:\:? \\ $$
Answered by A5T last updated on 07/May/24
$$\frac{\mathrm{1}}{{tan}\mathrm{75}}−{tan}\mathrm{75}=\frac{{cos}\mathrm{75}}{{sin}\mathrm{75}}−\frac{{sin}\mathrm{75}}{{cos}\mathrm{75}}=\frac{{cos}^{\mathrm{2}} \mathrm{75}−{sin}^{\mathrm{2}} \mathrm{75}}{{sin}\mathrm{75}{cos}\mathrm{75}} \\ $$$$=\frac{\mathrm{2}{cos}\mathrm{150}}{{sin}\mathrm{150}}=\frac{\mathrm{2}\left(\frac{−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}{\frac{\mathrm{1}}{\mathrm{2}}}=−\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow?=\frac{\mathrm{5}{sin}\left(\frac{\pi}{\mathrm{6}}\right)}{−\mathrm{2}\sqrt{\mathrm{3}}}=\frac{\frac{\mathrm{5}}{\mathrm{2}}}{−\mathrm{2}\sqrt{\mathrm{3}}}=\frac{−\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{12}} \\ $$