Question Number 207876 by hardmath last updated on 29/May/24
$$\mathrm{Find}:\:\:\:\mathrm{ln}\:\left(\frac{\mathrm{2}\:\mathrm{tg}\:\mathrm{22},\mathrm{30}°}{\mathrm{1}\:−\:\mathrm{tg}^{\mathrm{2}} \:\mathrm{22},\mathrm{30}°}\right)\:=\:? \\ $$
Commented by mr W last updated on 29/May/24
$${use}\:{your}\:{calculator}! \\ $$
Answered by som(math1967) last updated on 29/May/24
$${ln}\left({tan}\mathrm{2}×\mathrm{22}°\mathrm{30}'\right) \\ $$$$={lntan}\mathrm{45}={ln}\mathrm{1}=\mathrm{0} \\ $$
Commented by hardmath last updated on 29/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{professors} \\ $$
Commented by mr W last updated on 29/May/24
$${can}\:{you}\:{make}\:{your}\:{things}\:{clear} \\ $$$${and}\:{you}\:{should}\:{not}\:{let}\:{other}\:{people} \\ $$$${guess}.\:{if}\:{you}\:{mean}\:\mathrm{22}°\mathrm{30}'\:{then}\:{you} \\ $$$${should}\:{not}\:{write}\:\mathrm{22},\mathrm{30}°.\:{if}\:{you}\:{made} \\ $$$${typo},\:{you}\:{can}\:{correct}\:{it}. \\ $$
Commented by hardmath last updated on 29/May/24
$$\mathrm{dear}\:\mathrm{professor},\:\mathrm{yes},\:\mathrm{sorry}… \\ $$$$\mathrm{22},\mathrm{30}° \\ $$
Commented by mr W last updated on 29/May/24
$${then}\:{the}\:{result}\:{is}\:{not}\:\mathrm{0},\:{but} \\ $$$$\mathrm{ln}\:\left(\mathrm{tan}\:\mathrm{44}.\mathrm{6}°\right)\:{for}\:{which}\:{you}\:{need} \\ $$$${a}\:{calculator}. \\ $$
Commented by hardmath last updated on 29/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$