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..?


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Question Number 43847 by Rauny last updated on 16/Sep/18

$I think \tan 90 ° = 1 + i .$ $∵ \tan x = \frac{\sin x}{\cos x}$ $= \frac{e^{i x} - e^{- i x}}{2 i} \cdot \frac{2}{e^{i x} + e^{- i x}}$ $e^{i x} := E,$ $\frac{\sin x}{\cos x} = \frac{E - E^{- 1}}{(E + E^{- 1}) i} = - \frac{E - E^{- 1}}{E + E^{- 1}} i$ $= - \frac{(E - E^{- 1}) E}{(E + E^{- 1}) E} i = - \frac{E^{2} - 1}{E^{2} + 1} i$ $= - \frac{(E + 1) (E - 1)}{(E + i) (E - i)} i = - \frac{(E + 1) (E - 1)}{(E - i) (E + i)} i$ $= - \frac{E + 1}{E - i} \cdot \frac{E - 1}{E + i} i = - \frac{(E + 1) (E + i)}{(E - i) (E + i)} \cdot \frac{E - 1}{E + i} i$ $= - \frac{(E + 1) (E + i)}{E + 1} \cdot \frac{E - 1}{E + i} i = - (E + i) \cdot \frac{E - 1}{E + i} i$ $= - (E - 1) i$ $= - i (e^{i x} - 1) = \tan x$ $∴ \tan 90 ° = \tan \frac{π}{2} = - i (e^{\frac{1}{2} i π} - 1)$ $= - i (i - 1) = (- i) \times i - (- i) \times 1$ $= i + 1$ $= 1 + i$ $Right . . ?$
Commented by MrW3 last updated on 16/Sep/18

$\tan x = \frac{\sin x}{\cos x} i f \cos x \neq 0, i . e . x \neq n π + \frac{π}{2}$
Commented by maxmathsup by imad last updated on 16/Sep/18

$i h a v e u s e d t h e l i m i t . . .$
Commented by sma3l2996 last updated on 18/Sep/18

${Y o u}^{'} v e m a d e a m i s t a k e o n l i n e 9$ $\frac{(E + 1) (E + i)}{(E - i) (E + i)} = \frac{(E + 1) (E + i)}{E^{2} + 1} a n d d o e s n o t e q u a l$ $\frac{(E + 1) (E + i)}{E + 1}$
Commented by Rauny last updated on 19/Sep/18

$oh thx!!$ $finally I find the mistake by you .$ $thank you so much .$
Commented by sma3l2996 last updated on 19/Sep/18

$Y o u a r e w e l c o m e$
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