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Prove identity: $$\frac{1+\sin\alpha-\cos\alpha}{1+\sin\alpha+\cos\alpha}=\tan\frac{\alpha}{2}.$$ My work this far: we take the left side $$\dfrac{1+\sqrt{\frac{1-\cos2\alpha}{2}}-\sqrt{\frac{1+\cos2\alpha}{2}}}{1+\sqrt{\frac{1-\cos2\alpha}{2}}+\sqrt{\frac{1+\cos2\alpha}{2}}}=$$ then after some calulations I come to this $$=\frac{\sqrt{2} +\sqrt{1-\cos2\alpha}-\sqrt{1+\cos2\alpha}}{\sqrt{2} +\sqrt{1-\cos2\alpha}+\sqrt{1+\cos2\alpha}}$$ but now I'm stuck...

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    $\begingroup$ You transformed in the wrong direction. Express $\sin\alpha,\; \cos\alpha$ in terms of $\sin \frac{\alpha}{2}$ and $\cos \frac{\alpha}{2}$, not $\sin (2\alpha),\; \cos (2\alpha)$. $\endgroup$ – Daniel Fischer Aug 7 '15 at 12:23
  • $\begingroup$ I think that @ParthKohli provided a better answer. $\endgroup$ – Start wearing purple Aug 7 '15 at 13:37
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Express both $\sin\alpha$ and $\cos\alpha$ in terms of $t=\tan\frac{\alpha}{2}$: \begin{align*} \sin\alpha&=\frac{2t}{1+t^2},\\ \cos\alpha&=\frac{1-t^2}{1+t^2}. \end{align*}

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Write

  • $1 + \cos \alpha = 2\cos^2 \left(\frac{\alpha}{2}\right)$

  • $1 - \cos \alpha = 2 \sin^2 \left(\frac{\alpha}{2}\right)$

  • $\sin \alpha = 2 \sin \left(\frac{\alpha}{2}\right)\cos \left(\frac{\alpha }{2}\right)$

Things cancel out.

All of these are some or the other forms of the double-angle-identities.

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$\bf{My\; Solution::}$ Given $\displaystyle \frac{1+\sin \alpha-\cos \alpha}{1+\sin \alpha+\cos \alpha} = \frac{\left(1-\cos \alpha\right)+\sin \alpha}{\left(1+\cos \alpha\right)+\sin \alpha}$

$\displaystyle = \frac{2\sin^2 \frac{\alpha}{2}+2\sin \frac{\alpha}{2}\cdot \cos \frac{\alpha}{2}}{2\cos^2 \frac{\alpha}{2}+2\sin \frac{\alpha}{2}\cdot \cos \frac{\alpha}{2}} = \tan \frac{\alpha}{2}$

Above We have used the formula

$\star 1 + \cos \alpha = 2\cos^2 \left(\frac{\alpha}{2}\right)$

$\star 1 - \cos \alpha = 2 \sin^2 \left(\frac{\alpha}{2}\right)$

$\star \sin \alpha = 2 \sin \left(\frac{\alpha}{2}\right)\cos \left(\frac{\alpha }{2}\right)$

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Notice, $$LHS=\frac{1+\sin\alpha-\cos\alpha}{1+\sin\alpha+\cos\alpha}$$ $$=\frac{1+\frac{2\tan\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}-\frac{1-\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}}{1+\frac{2\tan\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}+\frac{1-\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}}$$ $$=\frac{1+\tan^2\frac{\alpha}{2}+2\tan\frac{\alpha}{2}-1+\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}+2\tan\frac{\alpha}{2}+1-\tan^2\frac{\alpha}{2}}$$

$$=\frac{2\tan\frac{\alpha}{2}+2\tan^2\frac{\alpha}{2}}{2+2\tan\frac{\alpha}{2}}$$

$$=\tan\frac{\alpha}{2}\left(\frac{1+\tan\frac{\alpha}{2}}{1+\tan\frac{\alpha}{2}}\right)$$$$=\tan\frac{\alpha}{2}=RHS$$

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Hint : Did you use the identity of $1 - cosα= 2sin^2a/2$ and $1 + cosα= 2cos^2a/2$

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