If-xy-e-4-Find-tg-ln-x-3-y-tg-ln-y-3-x-

Question Number 223179 by hardmath last updated on 16/Jul/25

$$\mathrm{If}:\:\:\:\mathrm{xy}\:=\:\boldsymbol{\mathrm{e}}^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{tg}\:\left(\mathrm{ln}\:\left(\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{y}}\right)\right)\:\centerdot\:\mathrm{tg}\:\left(\mathrm{ln}\:\left(\frac{\mathrm{y}^{\mathrm{3}} }{\mathrm{x}}\right)\right)\:=\:? \\ $$

Commented by hardmath last updated on 16/Jul/25

$$\mathrm{no}\:\mathrm{dear}\:\mathrm{professor},\:\mathrm{answer}: \\ $$$$\left.\mathrm{a}\left.\right)\left.\mathrm{1}\left.\:\left.\:\:\:\:\mathrm{b}\right)\frac{\mathrm{1}}{\mathrm{e}}\:\:\:\:\:\mathrm{c}\right)\frac{\mathrm{1}}{\pi}\:\:\:\:\:\mathrm{d}\right)\frac{\mathrm{e}}{\pi}\:\:\:\:\:\mathrm{e}\right)\frac{\pi}{\mathrm{e}} \\ $$

Answered by mr W last updated on 16/Jul/25

$${let}\:{X}=\mathrm{ln}\:{x},\:{Y}=\mathrm{ln}\:{y} \\ $$$$\mathrm{ln}\:\left({xy}\right)={X}+{Y}=\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{tan}\:\left(\mathrm{ln}\:\left(\frac{{x}^{\mathrm{3}} }{{y}}\right)\right)\centerdot\mathrm{tan}\:\left(\mathrm{ln}\:\left(\frac{{y}^{\mathrm{3}} }{{x}}\right)\right) \\ $$$$=\mathrm{tan}\:\left(\mathrm{3}{X}−{Y}\right)\centerdot\mathrm{tan}\:\left(\mathrm{3}{Y}−{X}\right) \\ $$$$=\mathrm{tan}\:\left(\mathrm{4}{X}−{X}−{Y}\right)\centerdot\mathrm{tan}\:\left(\mathrm{3}{Y}+\mathrm{3}{X}−\mathrm{4}{X}\right) \\ $$$$=\mathrm{tan}\:\left(\mathrm{4}{X}−\frac{\pi}{\mathrm{4}}\right)\centerdot\mathrm{tan}\:\left(\frac{\mathrm{3}\pi}{\mathrm{4}}−\mathrm{4}{X}\right) \\ $$$$=\mathrm{tan}\:\left(\mathrm{4}{X}−\frac{\pi}{\mathrm{4}}\right)\centerdot\mathrm{tan}\:\left(\frac{\pi}{\mathrm{2}}−\left(\mathrm{4}{X}−\frac{\pi}{\mathrm{4}}\right)\right) \\ $$$$=\mathrm{tan}\:\left(\mathrm{4}{X}−\frac{\pi}{\mathrm{4}}\right)\centerdot\mathrm{cot}\:\left(\mathrm{4}{X}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$=\mathrm{1}\:\checkmark \\ $$

Commented by mehdee7396 last updated on 16/Jul/25

$${in}\:{line}\:\mathrm{6} \\ $$$$\Rightarrow{tan}\left(\mathrm{4}{X}−\frac{\pi}{\mathrm{4}}\right)×{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}−\mathrm{4}{X}\right) \\ $$$$\Rightarrow\frac{{tan}\mathrm{4}{X}−\mathrm{1}}{\mathrm{1}+{tan}\mathrm{4}{X}}×\frac{−\mathrm{1}−{tan}\mathrm{4}{X}}{\mathrm{1}−{tan}\mathrm{4}{X}}=\mathrm{1} \\ $$$$ \\ $$

Commented by hardmath last updated on 16/Jul/25

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$

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