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Question Number 187330 by Humble last updated on 16/Feb/23

s h o w t h a t \frac{1 - c o s θ}{1 + c o s θ} = {t a n}^{2} (\frac{1}{2} θ)

Answered by Frix last updated on 16/Feb/23

What are we allowed to use ?

\tan^{2} x = \frac{\sin^{2} x}{\cos^{2} x} = \frac{\frac{1 - \cos 2 x}{2}}{\frac{1 + \cos 2 x}{2}} = \frac{1 - \cos 2 x}{1 + \cos 2 x}

\Rightarrow

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

or

\cos θ = \frac{e^{i θ} + e^{- i θ}}{2} = \frac{1 + e^{2 i θ}}{{2 e}^{i θ}}

\sin θ = \frac{1 - e^{2 i θ}}{{2 e}^{i θ}} i

\frac{1 - \cos θ}{1 + \cos θ} = \frac{- \frac{{(1 - e^{i θ})}^{2}}{{2 e}^{i θ}}}{\frac{{(1 + e^{i θ})}^{2}}{{2 e}^{i θ}}} = - {(\frac{1 - e^{i θ}}{1 + e^{i θ}})}^{2} =

= {(\frac{1 - e^{i θ}}{1 + e^{i θ}} i)}^{2} = {(\frac{1 - e^{2 i \frac{θ}{2}}}{1 + e^{2 i \frac{θ}{2}}} i)}^{2} = {(\frac{\frac{1 - e^{2 i \frac{θ}{2}}}{{2 e}^{i \frac{θ}{2}}} i}{\frac{1 + e^{2 i \frac{θ}{2}}}{{2 e}^{i \frac{θ}{2}}}})}^{2} =

= {(\frac{\sin \frac{θ}{2}}{\cos \frac{θ}{2}})}^{2} = \tan^{2} \frac{θ}{2}

Commented by Humble last updated on 16/Feb/23

a w e s o m e!! . t h a n k y o u, s i r

Commented by Frix last updated on 16/Feb/23

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