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Prove that $\dfrac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\left(\dfrac{\theta}{2}\right)$

Also it is a question of S.L. Loney's Plane Trignonometry

What I've tried by now:

\begin{align} & =\frac{1+\sin\theta-\sin(90-\theta)}{1+\cos(90-\theta)+\cos\theta} \\[10pt] & =\frac{1+2\cos45^\circ \sin(\theta-45^\circ)}{1+2\cos45^\circ \cos(45-\theta)} \end{align}

Cause I do know
\begin{align} & \sin c + \sin d = 2\sin\left(\frac{c+d}{2}\right)\cos\left(\frac{c-d}{2}\right) \\[10pt] \text{and } & \cos c + \cos d = 2\cos\left(\frac{c+d}{2}\right)\sin\left(\frac{c-d}{2}\right) \end{align}

I can't think of what to do next..

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  • $\begingroup$ An idea: Let $\theta = 2\phi$ $\endgroup$
    – GFauxPas
    Commented Aug 6, 2017 at 17:01
  • $\begingroup$ I didn't get that, what you trying to say. @GFauxPas $\endgroup$
    – TheMathGuy
    Commented Aug 6, 2017 at 17:03
  • $\begingroup$ Since the right-hand side can be written as $\sin\theta/(1+\cos\theta)$, the question "Better proof for ..." is related (and reciprocated). See if you can find $\cot(\theta/2)$ in my trigonographic answer. $\endgroup$
    – Blue
    Commented Aug 6, 2017 at 20:38

5 Answers 5

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Let $\theta = 2 \phi$, then the thing to be proven is:

Prove that $$\frac{1 + \sin(2\phi) - \cos(2\phi)}{1 + \sin(2\phi) + \cos(2\phi)} = \tan(\phi)$$

Then use:

$$\sin(2\phi) = 2 \sin \phi \cos \phi$$ $$\cos(2\phi) = \cos^2 \phi - \sin^2 \phi$$

and:

$$\sin^2 \phi + \cos^2 \phi = 1$$

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  • $\begingroup$ Note - as I was going to post very similarly: with $2\alpha = \theta$ your numerator becomes $$(\cos^2\alpha +\sin^2 \alpha)+2\sin \alpha\cos\alpha-(\cos^2\alpha -\sin^2 \alpha)=2\sin\alpha(\sin \alpha+\cos \alpha)$$ and the denominator $$(\cos^2\alpha +\sin^2 \alpha)+2\sin \alpha\cos\alpha+(\cos^2\alpha -\sin^2 \alpha)=2\cos\alpha(\sin \alpha+\cos \alpha)$$ $\endgroup$ Commented Aug 6, 2017 at 17:14
  • $\begingroup$ @MarkBennet I didn't want to do the problem for him, just give him a hint :) $\endgroup$
    – GFauxPas
    Commented Aug 6, 2017 at 17:17
  • $\begingroup$ Well, I didn't finish it - but I do think this is the best way rather than going straight to the $t$ substitution some of the other answers use, and I think the key is to do something with the $1$ in numerator and denominator to get a homogeneous expression - which is indicated by the expected answer. $\endgroup$ Commented Aug 6, 2017 at 17:34
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hint just use

$$1+\cos (X)=2\cos^2 (\frac {X}{2}) $$

$$1-\cos (X)=2\sin^2 (\frac {X}{2}) $$

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HINT: use the tan-half angle substution $$\sin(x)=2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2}}}$$ $$\cos(x)={\frac {1- \left( \tan \left( x/2 \right) \right) ^{2}}{1+ \left( \tan \left( x/2 \right) \right) ^{2}}} $$ with $t=\tan(x/2)$

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Write $t=\tan\frac{\theta}{2}$ so $\sin\theta=\frac{2t}{1+t^2},\,\cos\theta=\frac{1-t^2}{1+t^2}$. Hence $$\frac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}=\frac{1+t^2+2t-1+t^2}{1+t^2+2t+1-t^2}=\frac{2t+2t^2}{2+2t}=t.$$

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$$ \begin{aligned} & \frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}\\=& \frac{(1-\cos \theta)+\sin \theta}{(1+\cos \theta)+\sin \theta}\\ =& \frac{2 \sin ^{2}\left(\frac{\theta}{2}\right)+2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}{2 \cos ^{2}\left(\frac{\theta}{2}\right)+2 \sin \left(\frac{\theta}{2}\right) \operatorname{cor}\left(\frac{\theta}{2}\right)} \\ =& \frac{\sin \left(\frac{\theta}{2}\right)}{\cos \left(\frac{\theta}{2}\right)} \cdot \frac{\sin \left(\frac{\theta}{2}\right)+\cos \left(\frac{\theta}{2}\right)}{\cos \left(\frac{\theta}{2}\right)+\sin \left(\frac{\theta}{2}\right)} \\ =& \tan \left(\frac{\theta}{2}\right) \end{aligned} $$

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