Question Number 227286 by hardmath last updated on 12/Jan/26
$$\mathrm{tg}\left(\mathrm{4}\right)\:+\:\mathrm{ctg}\left(\mathrm{4}\right)\:=\:? \\ $$
Answered by Frix last updated on 12/Jan/26
$$=\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{8}} \\ $$
Answered by Kassista last updated on 13/Jan/26
$$ \\ $$$${tg}\left(\mathrm{4}\right)+\frac{\mathrm{1}}{{tg}\left(\mathrm{4}\right)}=\frac{{tg}^{\mathrm{2}} \left(\mathrm{4}\right)+\mathrm{1}}{{tg}\left(\mathrm{4}\right)}={sec}^{\mathrm{2}} \left(\mathrm{4}\right){cot}\left(\mathrm{4}\right) \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left(\mathrm{4}\right)}.\frac{{cos}\left(\mathrm{4}\right)}{{sin}\left(\mathrm{4}\right)}\:=\:\frac{\mathrm{1}}{{sin}\left(\mathrm{4}\right){cos}\left(\mathrm{4}\right)}\:=\:\frac{\mathrm{1}}{\frac{{sin}\left(\mathrm{8}\right)}{\mathrm{2}}} \\ $$$$\therefore\:=\:\frac{\mathrm{2}}{{sin}\left(\mathrm{8}\right)} \\ $$