Question Number 160411 by nadovic last updated on 29/Nov/21
$$\:\:\mathrm{Show}\:\mathrm{that}\:\:\mathrm{tan}\:\mathrm{58}°\mathrm{tan}\:\mathrm{32}°\:=\:\mathrm{1} \\ $$
Answered by TheSupreme last updated on 29/Nov/21
$${tan}\left(\alpha\right){tan}\left(\frac{\pi}{\mathrm{2}}−\alpha\right)={tan}\left(\alpha\right){cotan}\left(\alpha\right)=\mathrm{1} \\ $$$$ \\ $$