Question Number 154262 by mathlove last updated on 16/Sep/21
Answered by som(math1967) last updated on 16/Sep/21
$${lntan}\left(\frac{\pi}{\mathrm{180}}\right)+{lntan}\left(\frac{\mathrm{2}\pi}{\mathrm{180}}\right)+…{lntan}\left(\frac{\mathrm{89}\pi}{\mathrm{180}}\right) \\ $$$${lntan}\mathrm{1}°+{lntan}\mathrm{2}°+…{lntan}\mathrm{89}°\bigstar \\ $$$${lntan}\mathrm{1}×{tan}\mathrm{2}×…{tan}\mathrm{89} \\ $$$${lntan}\mathrm{1}×{cot}\left(\mathrm{90}−\mathrm{89}\right)×{tan}\mathrm{2}×{cot}\left(\mathrm{90}−\mathrm{88}\right)…×{tan}\mathrm{45} \\ $$$${lntan}\mathrm{1}×{cot}\mathrm{1}×{tan}\mathrm{2}×{cot}\mathrm{2}×…×\mathrm{1} \\ $$$$={ln}\mathrm{1}=\mathrm{0}\:{ans} \\ $$$$\bigstar\frac{\pi}{\mathrm{180}}=\mathrm{1}° \\ $$