Question

If $\cos \theta =\frac{\cos \alpha +\cos \beta }{1+\cos \alpha \cos \beta }$, prove that $\tan \frac{\theta }{2}=\pm \tan \frac{\alpha }{2}\tan \frac{\beta }{2}$

Answer 1

Given:

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

$$\begin{gathered} \Rightarrow \frac{1-{\tan }^2{\displaystyle \frac{\mathrm{\theta }}{2}}}{1+{\tan }^2\frac{\mathrm{\theta }}{2}}=\frac{\mathrm{cos\alpha }+\mathrm{cos\beta }}{1+\mathrm{cos\alpha }\times \mathrm{cos\beta }}\left[\because \mathrm{cos\theta }=\frac{1-{\tan }^2{\displaystyle \frac{\mathrm{\theta }}{2}}}{1+{\tan }^2{\displaystyle \frac{\mathrm{\theta }}{2}}}\right] \\ \mathrm{By}\mathrm{componendo}\mathrm{and}\mathrm{dividendo},\mathrm{we}\mathrm{get} \\ \frac{\left(1-{\tan }^2{\displaystyle \frac{\mathrm{\theta }}{2}}\right){\displaystyle +\left(1+{\tan }^2\frac{\mathrm{\theta }}{2}\right)}}{\left(1-{\tan }^2{\displaystyle \frac{\mathrm{\theta }}{2}}\right)-\left(1+{\tan }^2{\displaystyle \frac{\mathrm{\theta }}{2}}\right)}=\frac{\left(1+\mathrm{cos\alpha }\times \mathrm{cos\beta }+\mathrm{cos\alpha }+\mathrm{cos\beta }\right)}{-\left(1+\mathrm{cos\alpha cos\beta }-\mathrm{cos\alpha }-\mathrm{cos\beta }\right)} \\ \Rightarrow \frac{2}{2{\tan }^2{\displaystyle \frac{\mathrm{\theta }}{2}}}=\frac{\left(1+\mathrm{cos\alpha }\right)\left(1+\mathrm{cos\beta }\right)}{\left(1-\mathrm{cos\alpha }\right)\left(1-\mathrm{cos\beta }\right)} \end{gathered}$$

$$\begin{gathered} \Rightarrow {\tan }^2\frac{\mathrm{\theta }}{2}=\frac{\left(1-\mathrm{cos\alpha }\right)\left(1-\mathrm{cos\beta }\right)}{\left(1+\mathrm{cos\alpha }\right)\left(1+\mathrm{cos\beta }\right)} \\ \Rightarrow {\tan }^2\frac{\mathrm{\theta }}{2}=\frac{2{\sin }^2{\displaystyle \frac{\mathrm{\alpha }}{2}}\times 2{\sin }^2{\displaystyle \frac{\mathrm{\beta }}{2}}}{2{\cos }^2{\displaystyle \frac{\mathrm{\alpha }}{2}}\times 2{\cos }^2{\displaystyle \frac{\mathrm{\beta }}{2}}} \\ \Rightarrow {\tan }^2\frac{\mathrm{\theta }}{2}={\tan }^2\frac{\mathrm{\alpha }}{2}\times {\tan }^2\frac{\mathrm{\beta }}{2} \\ \Rightarrow \tan \frac{\mathrm{\theta }}{2}=\pm \tan \frac{\mathrm{\alpha }}{2}\times \tan \frac{\mathrm{\beta }}{2} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$