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So, for any angle $\alpha$ : $$\cos(2\alpha) = \cos^2\alpha - \sin^2\alpha = \dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha+\sin^2\alpha} = \dfrac{\dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha}}{\dfrac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}}= \dfrac{1-\tan^2\alpha}{1+\tan^2\alpha}$$ Now, $\cos\alpha = \cos\Big(2\cdot\dfrac{\alpha}{2}\Big) = \dfrac{1-\tan^2\dfrac{\alpha}{2}}{1+\tan^2\dfrac{\alpha}{2}}$

Now, using the componendo and dividendo rule, we get : $$\dfrac{\cos\alpha+1}{\cos\alpha-1} = \dfrac{2}{-2\tan^2\dfrac{\alpha}{2}} = \dfrac{-1}{\tan^2\dfrac{\alpha}{2}} \implies \tan^2\dfrac{\alpha}{2} = \dfrac{1-\cos\alpha}{1+\cos\alpha}$$ $$\implies \tan^2\dfrac{\alpha}{2} = \dfrac{(1-\cos\alpha)(1-\cos\alpha)}{(1+\cos\alpha)(1-\cos\alpha)} = \Big(\dfrac{1-\cos\alpha}{\sin\alpha}\Big)^2$$ $$\implies \Bigg|\tan\Big(\dfrac{\alpha}{2}\Big)\Bigg| = \Bigg|\dfrac{1-\cos\alpha}{\sin\alpha}\Bigg|$$ Now, only if $\mathrm{sign}\Big(\tan\dfrac{\alpha}{2}\Big) = \mathrm{sign}\Big(\dfrac{1-\cos\alpha}{\sin\alpha}\Big)$ is true, we can say that $\tan\dfrac{\alpha}{2} = \dfrac{1-\cos\alpha}{\sin\alpha}$

So, I think that without proving that, the proof will be incomplete but my Math textbook doesn't prove it.

So, is it necessary to prove it? If not, why not?

Thanks!

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Your comment is correct. You can only get the final equality by proving that $\tan\left(\frac{\alpha}{2}\right)$ and $\frac{1-\cos \alpha}{\sin\alpha}$ have the same sign.

But this is not complicated to prove. $\tan\left(\frac{\alpha}{2}\right)$ is positive if and only if $\frac{\alpha}{2} \in (k\pi, k\pi +\frac{\pi}{2})$. Like $\sin \alpha$ while $1- \cos \alpha$ is always non negative.

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Here's a simple way of proving the identity: $$\frac{1-\cos x}{\sin x}=\frac{1-(1-2\sin^2\frac{x}{2})}{2\sin\frac{x}{2}\cos\frac{x}{2}}=\frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=\tan\frac{x}{2} $$ as required. I hope that was helpful:)

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    $\begingroup$ $+1$. Nice way to derive it. I didn't accept your answer because that wasn't exactly what my question asked but a great answer, nonetheless :-) $\endgroup$ – Rajdeep Sindhu Aug 25 '20 at 15:43
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    $\begingroup$ @RajdeepSindhu thanks for the compliment, it;s a pleasure to help! :) You might also be interested in the following: $$\frac{\sin2x}{1+\cos2x}=\frac{1-\cos2x}{\sin 2x}=\tan x$$ which can be proved in a very similar way. Don't worry about not accepting the answer,it's helping you that counts :) $\endgroup$ – A-Level Student Aug 25 '20 at 16:53
  • $\begingroup$ It's helping you that counts. People like you are the reason I love the Stack Exchange community. :) $\endgroup$ – Rajdeep Sindhu Aug 26 '20 at 16:56
  • $\begingroup$ @RajdeepSindhu thank you for the compliment, I appreciate it :)) $\endgroup$ – A-Level Student Aug 27 '20 at 18:33
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Yes, it is necessary to show.

As $\tan$ has a periodicity $\pi$, it's enough to check the signs for $\dfrac{\alpha}{2}$ in the ranges, $\left[0,\dfrac{\pi}{4}\right], \left[\dfrac{\pi}{4},\dfrac{\pi}{2}\right),\left(\dfrac{\pi}{2},\dfrac{3\pi}{4}\right] $ and $\left[\dfrac{3\pi}{4},\pi\right]$

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