# How to prove that $\tan^2(\frac\theta2)= \tan^2(\frac\alpha2)\tan^2(\frac\beta2)$?

I'm unable to solve this question:
$$\cos(\theta)=\dfrac{\cos(\alpha)+\cos(\beta)}{1+\cos(\alpha) \cos(\beta)}$$

Prove: $$\tan^2\left(\frac\theta2\right)= \tan^2\left(\frac\alpha2\right)\tan^2\left(\frac\beta2\right)$$

I have tried the following:

1. Using the indentity: $$\cos(\theta)=\dfrac{1-\tan^2\left(\frac\theta2\right)}{1+\tan^2\left(\frac\theta2\right)}$$
2. Diving by $$\cos (\alpha) \cos (\beta)$$
3. Creating a triangle, to find $$\tan(\theta)= \sin \alpha \sin \beta$$

Every time I got a huge complex equation with roots. Any help would be appreciated.

• In your step 3, I think it should be $\tan( \theta) = \dfrac {\sin \alpha \sin \beta}{\cos(\alpha) + \cos (\beta)}$ instead. – Mick Aug 18 '14 at 15:07
• Oh sorry, it is. :) – Harshal Gajjar Aug 18 '14 at 15:51
• – lab bhattacharjee Aug 18 '14 at 16:35

$$\tan ^2 {\frac {\theta}{2}} = \frac{\sin^2 {\frac {\theta}{2}} }{\cos^2 {\frac {\theta}{2}}} = \frac{ \left({\dfrac{1 - \cos \theta}{2}}\right)}{\left({\dfrac{1 + \cos \theta}{2}}\right)} \;\; \text{(since \cos 2 \theta = \cos^2 \theta - \sin^2 \theta= 1 - 2 \sin^2 \theta = 2 \cos^2\theta - 1 )}$$

Now substituting for $\cos \theta$ and factorising we have,

$$= \frac{1 + \cos \alpha \cos \beta - \cos \alpha - \cos \beta }{ 1 + \cos \alpha \cos \beta + \cos \alpha + \cos \beta} = \frac{(1 - \cos \alpha)(1 - \cos \beta)}{(1 + \cos \alpha)(1 + \cos \beta)}$$

Now use the same formulas for $cos 2 \theta$ mentioned above strategically to negate the $1$ present in all the factors and to introduce $\dfrac {\alpha}{2}$. Then,

$$= \frac{[1 - (1 - 2 \sin^2 \frac {\alpha}{2} ) ][1 - ( 1 - 2 \sin^2 \frac {\beta}{2})]}{[1 + (2 \cos^2 \frac {\alpha}{2} - 1)][1 + (2 \cos ^2 \frac {\beta}{2} - 1)]}$$

Remember $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$.

Now Simplify!!

You may try componendo and dividendo :

$\dfrac{1-\tan^2\left(\frac\theta2\right)}{1+\tan^2\left(\frac\theta2\right)}=\dfrac{\cos(\alpha)+cos(\beta)}{1+\cos(\alpha) \cos(\beta)} \\ \iff \dfrac{-2\tan^2\left(\frac\theta2\right)}{2} \stackrel{\color{red}{*}}{=} \dfrac{\cos(\alpha)+cos(\beta) - 1-\cos(\alpha)\cos(\beta)}{\cos(\alpha)+cos(\beta) + 1+\cos(\alpha)\cos(\beta)} \\ \iff \tan^2\left(\frac\theta2\right) = \dfrac{(1-\cos(\alpha))(1-\cos(\beta))}{(1+\cos(\alpha))(1+\cos(\beta))}$

$\color{Red}{*} :$ componendo and dividendo

• Always eager to learn something new. Thanks! – Han de Bruijn Aug 18 '14 at 15:03
• np :) c&d is one of those beautiful algebra methods which is something i have never seen mentioned in any highschool text books! – AgentS Aug 18 '14 at 15:11