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# If $\cos \theta=\cos \alpha.\cos\beta$ prove that $\tan\large\frac{\theta+\alpha}{2}$$\tan\large\frac{\theta-\alpha}{2}=$$\tan^2\large\frac{\beta}{2}$

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A)
Toolbox:
• $\tan \large\frac{\theta}{2}=\frac{1-\cos \theta}{\sin \theta}$
• $\cos a-\cos b=-2\sin \large\frac{(a+b)}{2}$$\sin \large\frac{(a-b)}{2} • \cos a+\cos b=2\cos \large\frac{(a+b)}{2}$$\cos \large\frac{(a-b)}{2}$
$\cos \theta=\cos \alpha.\cos\beta$
$\therefore \cos \beta=\large\frac{\cos \theta}{\cos \alpha}$-----(1)
RHS
$\tan^2\large\frac{\beta}{2}=\frac{(1-\cos \beta)^2}{\sin ^2\beta}$
$\Rightarrow \large\frac{(1-\cos \beta)^2}{(1-\cos^2\beta)}=\frac{(1-\cos \beta)^2}{1-\cos\beta)(1+\cos \beta)}$
$\Rightarrow \large\frac{1-\cos \beta}{1+\cos \beta}=\frac{1-\Large\frac{\cos \theta}{\cos \alpha}}{1+\Large\frac{\cos \theta}{\cos \alpha}}$
$\Rightarrow \large\frac{\cos \alpha-\cos \theta}{\cos \alpha+\cos \theta}$
$\Rightarrow \large\frac{2\sin \large\frac{\alpha+\theta}{2}\sin \Large\frac{\theta-\alpha}{2}}{2\cos\Large\frac{\theta+\alpha}{2}\cos\Large\frac{\theta-\alpha}{2}}$
$\Rightarrow \tan \large\frac{\theta+\alpha}{2}.\frac{\theta-\alpha}{2}$
Hence proved