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Question Number 43847 by Rauny last updated on 16/Sep/18
I think tan 90°=1+i. ∵tan x=((sin x)/(cos x)) =((e^(ix) −e^(−ix) )/(2i))∙(2/(e^(ix) +e^(−ix) )) e^(ix) :=E, ((sin x)/(cos x))=((E−E^(−1) )/((E+E^(−1) )i))=−((E−E^(−1) )/(E+E^(−1) ))i =−(((E−E^(−1) )E)/((E+E^(−1) )E))i=−((E^2 −1)/(E^2 +1))i =−(((E+1)(E−1))/((E+i)(E−i)))i=−(((E+1)(E−1))/((E−i)(E+i)))i =−((E+1)/(E−i))∙((E−1)/(E+i))i=−(((E+1)(E+i))/((E−i)(E+i)))∙((E−1)/(E+i))i =−(((E+1)(E+i))/(E+1))∙((E−1)/(E+i))i=−(E+i)∙((E−1)/(E+i))i =−(E−1)i =−i(e^(ix) −1)=tan x ∴tan 90°=tan (π/2)=−i(e^((1/2)iπ) −1) =−i(i−1)=(−i)×i−(−i)×1 =i+1 =1+i Right..?
$$\mathrm{I}\:\mathrm{think}\:\mathrm{tan}\:\mathrm{90}°=\mathrm{1}+{i}. \\ $$$$\because\mathrm{tan}\:{x}=\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}} \\ $$$$=\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}\centerdot\frac{\mathrm{2}}{{e}^{{ix}} +{e}^{−{ix}} } \\ $$$${e}^{{ix}} :={E}, \\ $$$$\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}=\frac{{E}−{E}^{−\mathrm{1}} }{\left({E}+{E}^{−\mathrm{1}} \right){i}}=−\frac{{E}−{E}^{−\mathrm{1}} }{{E}+{E}^{−\mathrm{1}} }{i} \\ $$$$=−\frac{\left({E}−{E}^{−\mathrm{1}} \right){E}}{\left({E}+{E}^{−\mathrm{1}} \right){E}}{i}=−\frac{{E}^{\mathrm{2}} −\mathrm{1}}{{E}^{\mathrm{2}} +\mathrm{1}}{i} \\ $$$$=−\frac{\left({E}+\mathrm{1}\right)\left({E}−\mathrm{1}\right)}{\left({E}+{i}\right)\left({E}−{i}\right)}{i}=−\frac{\left({E}+\mathrm{1}\right)\left({E}−\mathrm{1}\right)}{\left({E}−{i}\right)\left({E}+{i}\right)}{i} \\ $$$$=−\frac{{E}+\mathrm{1}}{{E}−{i}}\centerdot\frac{{E}−\mathrm{1}}{{E}+{i}}{i}=−\frac{\left({E}+\mathrm{1}\right)\left({E}+{i}\right)}{\left({E}−{i}\right)\left({E}+{i}\right)}\centerdot\frac{{E}−\mathrm{1}}{{E}+{i}}{i} \\ $$$$=−\frac{\left({E}+\mathrm{1}\right)\left({E}+{i}\right)}{{E}+\mathrm{1}}\centerdot\frac{{E}−\mathrm{1}}{{E}+{i}}{i}=−\left({E}+{i}\right)\centerdot\frac{{E}−\mathrm{1}}{{E}+{i}}{i} \\ $$$$=−\left({E}−\mathrm{1}\right){i} \\ $$$$=−{i}\left({e}^{{ix}} −\mathrm{1}\right)=\mathrm{tan}\:{x} \\ $$$$\therefore\mathrm{tan}\:\mathrm{90}°=\mathrm{tan}\:\frac{\pi}{\mathrm{2}}=−{i}\left({e}^{\frac{\mathrm{1}}{\mathrm{2}}{i}\pi} −\mathrm{1}\right) \\ $$$$=−{i}\left({i}−\mathrm{1}\right)=\left(−{i}\right)×{i}−\left(−{i}\right)×\mathrm{1} \\ $$$$={i}+\mathrm{1} \\ $$$$=\mathrm{1}+{i} \\ $$$$\mathrm{Right}..? \\ $$
Commented by MrW3 last updated on 16/Sep/18
tan x=((sin x)/(cos x)) if cos x≠0, i.e. x≠nπ+(π/2)
$$\mathrm{tan}\:{x}=\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\:{if}\:\mathrm{cos}\:{x}\neq\mathrm{0},\:{i}.{e}.\:{x}\neq{n}\pi+\frac{\pi}{\mathrm{2}} \\ $$
Commented by maxmathsup by imad last updated on 16/Sep/18
i have used the limit ...
$${i}\:{have}\:{used}\:{the}\:{limit}\:… \\ $$
Commented by sma3l2996 last updated on 18/Sep/18
You′ve made a mistake on line 9 (((E+1)(E+i))/((E−i)(E+i)))=(((E+1)(E+i))/(E^2 +1)) and does not equal (((E+1)(E+i))/(E+1))
$${You}'{ve}\:{made}\:{a}\:{mistake}\:{on}\:{line}\:\mathrm{9} \\ $$$$\frac{\left({E}+\mathrm{1}\right)\left({E}+{i}\right)}{\left({E}−{i}\right)\left({E}+{i}\right)}=\frac{\left({E}+\mathrm{1}\right)\left({E}+{i}\right)}{{E}^{\mathrm{2}} +\mathrm{1}}\:{and}\:{does}\:{not}\:{equal} \\ $$$$\frac{\left({E}+\mathrm{1}\right)\left({E}+{i}\right)}{{E}+\mathrm{1}} \\ $$
Commented by Rauny last updated on 19/Sep/18
oh thx!! finally I find the mistake by you. thank you so much.
$$\mathrm{oh}\:\mathrm{thx}!! \\ $$$$\mathrm{finally}\:\mathrm{I}\:\mathrm{find}\:\mathrm{the}\:\mathrm{mistake}\:\mathrm{by}\:\mathrm{you}. \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}. \\ $$
Commented by sma3l2996 last updated on 19/Sep/18
You are welcome
$${You}\:{are}\:{welcome} \\ $$

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