For the first problem,
clearly, we need to eliminate $\theta,\phi$
Now, $\cos\phi=\frac{\cos\alpha}{\cos\beta}$ and $\cos\theta=\frac{\cos\alpha}{\cos\gamma}$
$$\text{and }\sin^2\alpha=\left(2\sin\dfrac\phi2\cdot\sin\dfrac\theta2\right)^2=(2\sin^2\dfrac\phi2)(2\sin^2\dfrac\theta2)=(1-\cos\phi)(1-\cos\theta)$$
$$\implies 1-\cos^2\alpha=\left(1-\frac{\cos\alpha}{\cos\beta}\right)\left(1-\frac{\cos\alpha}{\cos\gamma}\right)$$
$$\implies \cos^2\alpha(1+\cos\beta\cos\gamma)=\cos\alpha(\cos\beta+\cos\gamma)$$
Assuming $\cos\alpha\ne0, \cos\alpha=\frac{\cos\beta+\cos\gamma}{1+\cos\beta\cos\gamma}$
Applying Componendo and dividendo, $$\frac{1-\cos\alpha}{1+\cos\alpha}=\frac{1+\cos\beta\cos\gamma-(\cos\beta+\cos\gamma)}{1+\cos\beta\cos\gamma+\cos\beta+\cos\gamma}=\frac{(1-\cos\beta)(1-\cos\gamma)}{(1+\cos\beta)(1+\cos\gamma)}$$
As $\cos2A=\frac{1-\tan^2A}{1+\tan^2A}, \tan^2A=\frac{1-\cos2A}{1+\cos2A}$
$$\implies \tan^2\frac\alpha2=\tan^2\frac\beta2\cdot\tan^2\frac\gamma2 $$
For the second question, $$\tan\alpha= \tan^3\frac\theta2$$
$$\implies \frac{\sin\alpha}{\sin^3\frac\theta2}=\frac{\cos \alpha}{\cos^3\frac\theta2}=\pm\sqrt{\frac{\sin^2\alpha+\cos^2\alpha}{\left(\sin^3\frac\theta2\right)^2+\left(\cos^3\frac\theta2\right)^2}}$$
$$\text{Now,} \left(\sin^3\frac\theta2\right)^2+\left(\cos^3\frac\theta2\right)^2=\left(\sin^2\frac\theta2\right)^3+\left(\cos^2\frac\theta2\right)^3$$
$$=\left(\frac{1-\cos\theta}2\right)^3+\left(\frac{1+\cos\theta}2\right)^3 \text{ as }\cos2A=1-2\sin^2=2\cos^2A-1$$
$$=\frac{2(1+3\cos^2\theta)}8=\frac{m^2}4 \text{ as }\cos^2\theta=\frac{m^2-1}3$$
$$\implies \frac{\sin\alpha}{\sin^3\frac\theta2}=\frac{\cos \alpha}{\cos^3\frac\theta2}=\pm \frac2m$$
$$\implies \sin^3\frac\theta2=\pm\frac{m\sin\alpha}2\implies \sin\frac\theta2=\left(\pm\frac{m\sin\alpha}2\right)^{\frac13}$$
Similarly, find $\cos \frac\theta2$ and use $\cos^2\frac\theta2+\sin^2\frac\theta2=1$ to eliminate $\theta$
Best Answer
$\tan(\phi-\theta)=\frac{\tan\phi-\tan\theta}{1+\tan\phi\tan\theta}$.
If we replace $\tan\theta=\cos 2\alpha \tan \phi$ we get
\begin{align*} \frac{\tan\phi-\cos 2\alpha\tan\phi}{1+\tan\phi\cos2\alpha\tan\phi}&=\frac{\tan\phi-(\cos^2\alpha-\sin^2 \alpha)\tan\phi}{1+\tan^2\phi(\cos^2\alpha-\sin^2\alpha)}\\ &=\frac{2\sin^2\alpha\tan\phi\cos^2\phi}{\cos^2\phi+\sin^2\phi(\cos^2\alpha-\sin^2\alpha)}\\ &=\frac{2\sin^2\alpha \sin\phi\cos\phi}{\cos^2\phi+\sin^2\phi(1-2\sin^2\alpha)}\\ &=\frac{\sin^2\alpha \sin 2\phi}{1-2\sin^2\alpha\sin^2\phi}\\ &=\frac{\sin^2\alpha \sin 2\phi}{\cos^2\alpha+\sin^2\alpha-2\sin^2\alpha\sin^2\phi}\\ &=\frac{\sin^2\alpha\sin 2\phi}{\cos^2\alpha+\sin^2\alpha(1-2\sin^2\phi)}\\&=\frac{\sin^2\alpha\sin2\phi}{\cos^2\alpha(1+\tan^2\alpha \cos2\phi)}\\ &=\frac{\tan^2\alpha \sin 2\phi }{1+\tan^2\alpha \cos2\phi}. \end{align*}