# Prove trigonometric identity for $\tan^2(\theta/2)$

I've been at this for a couple of days (3-4 hours total now), and am feeling lost:

$$(1)\ \tan^2(\frac{1}{2}\theta) = \frac{\tan(\theta) - \sin(\theta)}{\tan(\theta)+\sin(\theta)}$$

I'm aware that $$(2)\ \tan^2(\frac{1}{2}\theta)=\frac{1 - \cos(\theta)}{1 + \cos(\theta)} = \frac{\left(\frac{1 - \cos(\theta)}{2}\right)}{\left(\frac{1 + \cos(\theta)}{2}\right)}$$

And have been working towards solving the right-hand-side of the first identity listed into something along the lines of the right-hand-side of the second identity...though, I've been having no success.

Any thoughts?

Edit

For the sake of clarification, I'm also aware of the double angle rules and how they are used to obtain $\tan^2(\frac{\theta}{2})$; my issue is understanding the connection between those rules and what exists on the RHS within equation $(1)$. Apologies for any confusion on this.

• From the second line, cancel the 2's in the denominator and multiply the top and bottom by $\mathrm{tan}(\theta)$. – Michael Kasa Feb 18 '14 at 21:01
• If you replace $\tan\theta$ on the right-hand side by $\frac{\sin\theta}{\cos\theta}$ and fool around a little, you will get something familiar. – André Nicolas Feb 18 '14 at 21:04
• @AndreNicolas: believe it or not, that's a large portion of what I've been trying. I always end up with something like $\frac{\sin(\theta) - \sin(\theta)\cos(\theta)}{\cos(\theta)}$ in the numerator, and $\frac{\sin(\theta) + \sin(\theta)\cos(\theta)}{\cos(\theta)}$ in the denominator. It's from there that I feel lost. – about blank Feb 18 '14 at 21:29
• @aboutblank: You're close! You have "stuff-over-cosine" in the numerator, and "otherstuff-over-cosine" in the denominator, so those cosines cancel. (Why?) This leaves "stuff-over-otherstuff". Factoring the "stuff" and the "otherstuff" will lead to more cancellation, and you're practically there! – Blue Feb 18 '14 at 21:44
• The first top is $\sin\theta(1-\cos\theta)$. The second top is $\sin\theta(1+\cos\theta)$. Divide. – André Nicolas Feb 18 '14 at 21:59

$$\frac{\tan(\theta) - \sin(\theta)}{\tan(\theta)+\sin(\theta)}=\frac{\sin(\theta)/\cos\theta - \sin(\theta)}{\sin(\theta)/\cos\theta+\sin(\theta)}=\frac{1/\cos\theta - 1}{1/\cos\theta+1}=\frac{1-\cos\theta }{1+\cos\theta}$$ You can use the double angle rule to get $$\frac{1 - \cos(\theta)}{1 + \cos(\theta)}=\frac{1 -1+ 2\sin^2(\theta/2)}{1 + 2\cos^2(\theta/2)-1}=\frac{\sin^2(\theta/2)}{\cos^2(\theta/2)}=\tan^2(\theta/2)$$