Question Number 185732 by Rupesh123 last updated on 26/Jan/23
Answered by witcher3 last updated on 26/Jan/23
$$\mathrm{tg}\left(\mathrm{a}\right)−\mathrm{tg}\left(\mathrm{b}\right)=\frac{\mathrm{sin}\left(\mathrm{a}−\mathrm{b}\right)}{\mathrm{cos}\left(\mathrm{a}\right)\mathrm{cos}\left(\mathrm{b}\right)} \\ $$$$\mathrm{tg}\left(\mathrm{50}\right)−\mathrm{tg}\left(\mathrm{40}\right)=\frac{\mathrm{sin}\left(\mathrm{10}\right)}{\mathrm{cos}\left(\mathrm{40}\right)\mathrm{cos}\left(\mathrm{50}\right)} \\ $$$$\Leftrightarrow\frac{\mathrm{cos}\left(\mathrm{10}\right)}{\mathrm{cos}\left(\mathrm{40}\right)\mathrm{cos}\left(\mathrm{50}\right)}=\frac{\mathrm{cos}\left(\mathrm{10}\right)}{\mathrm{cos}\left(\mathrm{40}\right)\mathrm{sin}\left(\mathrm{40}\right)}=\mathrm{2}\frac{\mathrm{cos}\left(\mathrm{10}\right)}{\mathrm{2cos}\left(\mathrm{40}\right)\mathrm{sin}\left(\mathrm{40}\right)} \\ $$$$=\frac{\mathrm{2cos}\left(\mathrm{10}\right)}{\mathrm{sin}\left(\mathrm{80}\right)}=\mathrm{2} \\ $$