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Question Number 73560 by mhmd last updated on 13/Nov/19

p r o v e t h a t (1 - i t a n θ) / (i + c o t θ) = i t a n θ

p l e a s s i r h e l p m e ?

Commented by MJS last updated on 13/Nov/19

{it}^{'} s not generally true

\frac{1 - i \tan θ}{i + \cot θ} = 2 \sin θ \cos θ - \tan θ - 2 i \sin^{2} θ

{we}^{'} d have to have

2 \sin θ \cos θ - \tan θ = 0 \land - {2 \sin}^{2} θ = \tan θ

which is only true for

θ = 2 n π \lor θ = \frac{8 n - 1}{4} π \lor θ = \frac{8 n + 3}{4} π

Answered by ajfour last updated on 13/Nov/19

\frac{1 - i \tan θ}{i + \cot θ} = \frac{(1 - i \tan θ) (\tan θ)}{1 + i \tan θ}

= {(1 - i \tan θ)}^{2} \sin θ \cos θ

= e^{- 2 i θ} \tan θ

. .

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