Question Number 107684 by bemath last updated on 12/Aug/20
$$\:\:\:\:\:“\mathcal{B}{e}\mathcal{M}{ath}“ \\ $$$$\:\mathrm{ln}\:\mathrm{tan}\:\mathrm{1}°+\mathrm{ln}\:\mathrm{tan}\:\mathrm{2}°+\mathrm{ln}\:\mathrm{tan}\:\mathrm{3}°+…+\mathrm{ln}\:\mathrm{tan}\:\mathrm{89}°=? \\ $$
Answered by Her_Majesty last updated on 12/Aug/20
$${ln}\:{tan}\:{x}°\:=−{ln}\:{tan}\:\left(\mathrm{90}−{x}\right)° \\ $$$$\Rightarrow\:{answer}\:{is}\:{ln}\:{tan}\:\mathrm{45}°\:=\mathrm{0} \\ $$
Commented by bemath last updated on 12/Aug/20
$${thank}\:{you}\:{Miss}\: \\ $$