# If $\cos\theta=\frac{\cos\alpha+\cos \beta}{1+\cos\alpha\cos\beta}$, then prove that one value of $\tan(\theta/2)$ is $\tan(\alpha/2)\tan(\beta/2)$

If $$\cos\theta = \frac{\cos\alpha + \cos \beta}{1 + \cos\alpha\cos\beta}$$ then prove that one of the values of $$\tan{\frac{\theta}{2}}$$ is $$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}$$.

I don't even know how to start this question. Pls help

• If you can write $\sin(\theta/2)$ and $\cos(\theta/2)$ in terms of $\cos\theta$, then you can write $\tan(\theta/2)$ in terms of $\cos\theta$. – Blue Feb 13 at 0:49
• Maybe this will help? $\tan(\theta/2)=\frac {1-\cos(\theta)}{\sin(\theta)}=\frac {\sin(\theta)}{1+\cos(\theta)}$ en.wikipedia.org/wiki/… – Akash Patel Feb 13 at 0:53
• quora.com/… – Dr. Mathva Feb 13 at 0:56
• – lab bhattacharjee Feb 13 at 2:24

$$\tan{\frac{\theta}{2}} = \pm \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} \longrightarrow \text{eq.1}$$ Evaluating $$\frac{1 - \cos\theta}{1 + \cos\theta}$$ first,
$$\frac{1 - \cos\theta}{1 + \cos\theta} = \frac{1 - (\frac{\cos\alpha + \cos\beta}{1 + \cos\alpha\cos\beta})}{1 + (\frac{\cos\alpha + \cos\beta}{1 + \cos\alpha\cos\beta})}\text{ }[\text{since } \cos\theta= \frac{\cos\alpha +\cos\beta}{1 + \cos\alpha\cos\beta}]$$ $$=\frac{1+\cos\alpha\cos\beta - \cos\alpha -\cos\beta}{1+\cos\alpha\cos\beta + \cos\alpha+ \cos\beta}$$ $$=\frac{(1-\cos\alpha)(1-\cos\beta)}{(1+\cos\alpha)(1+\cos\beta)}$$
Substituting this value of $$\frac{1 - \cos\theta}{1 + \cos\theta}$$ into equation 1,
$$\tan{\frac{\theta}{2}} = \sqrt{\frac{(1-\cos\alpha)(1-\cos\beta)}{(1+\cos\alpha)(1+\cos\beta)}}$$ $$=\sqrt{\frac{(1-\cos\alpha)^2(1-\cos\beta)^2}{(1-\cos^2\alpha)(1-\cos^2\beta)}}$$ $$=\pm\frac{(1-\cos\alpha)(1-\cos\beta)}{\sin\alpha\sin\beta}$$ Taking the positive value of $$\tan{\frac{\theta}{2}}$$, $$\frac{(1-\cos\alpha)(1-\cos\beta)}{\sin\alpha\sin\beta} = \frac{4\sin^2\frac{\alpha}{2}\sin^2\frac{\beta}{2}}{4\sin\frac{\alpha}{2}\cos{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}} \text{(using half angle and double angle formula)}$$ $$=\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}$$
Therefore, $$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}$$ is one of the values of $$\tan{\frac{\theta}{2}}$$.