If $$\cos\alpha=\cos\beta\cos\phi=\cos\gamma\cos\theta$$ and $$\sin\alpha=2\sin\frac{\phi}{2}\sin\frac{\theta}{2}$$ then prove that $$\tan\frac{\alpha}{2}=\pm\tan\frac{\beta}{2}\tan\frac{\gamma}{2}$$
My approach:
$$\frac{\cos \alpha}{\cos\beta}=\cos\phi$$ $$=1-2\sin^2\frac{\phi}{2}$$ $$\Rightarrow 2\sin^2\frac{\phi}{2}=1-\frac{\cos\alpha}{\cos\beta}$$ and(in the similar fashion) $$\Rightarrow 2\sin^2\frac{\theta}{2}=1-\frac{\cos\alpha}{\cos\gamma}$$
Please help me to proceed.
$$\sin^2\alpha=2\sin^2\dfrac\phi2\cdot2\sin^2\dfrac\theta2$$
$$=\left(1-\frac{\cos\alpha}{\cos\beta}\right)\left(1-\frac{\cos\alpha}{\cos\gamma}\right)$$
$$\cos\beta\cos\gamma(1-\cos^2\alpha)=(\cos\beta-\cos\alpha)(\cos\gamma-\cos\alpha)$$
$$-\cos^2\alpha\cos\beta\cos\gamma=-\cos\alpha\cos\beta-\cos\alpha\cos\gamma+\cos^2\alpha$$
Assuming $\cos\alpha\ne0,$
$$-\cos\alpha\cos\beta\cos\gamma=-\cos\beta-\cos\gamma+\cos\alpha$$
$$\iff\cos\alpha(1+\cos\beta\cos\gamma)=\cos\beta+\cos\gamma$$
$$\dfrac{\cos\beta+\cos\gamma}{1+\cos\beta\cos\gamma}=\dfrac{\cos\alpha}1$$
Apply Componendo and Dividendo and use $$\cos2x=\dfrac{1-\tan^2x}{1+\tan^2x}$$