Question

If $\cos \theta =\frac{\cos u-e}{1-e\cos u},$ then show that $\tan \frac{\theta }{2}=\pm \sqrt{\frac{1+e}{1-e}}\tan \left(\frac{u}{2}\right)$.

Answer 1

$\cos \theta =\frac{\cos u-e}{1-e\cos u}$

using trigonometric identities and simplifying

$\Rightarrow \frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}=\frac{\frac{1-{\tan }^2\frac{u}{2}}{1+{\tan }^2\frac{u}{2}}-e}{1-e\left(\frac{1-{\tan }^2\frac{u}{2}}{1+{\tan }^2\frac{u}{2}}\right)}$
$\Rightarrow \frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}=\frac{1-{\tan }^2\frac{u}{2}-e(1+{\tan }^2\frac{u}{2})}{1+{\tan }^2\frac{u}{2}-e(1-{\tan }^2\frac{u}{2})}$

Applying componendo and dividendo, we get

$\Rightarrow \frac{1-{\tan }^2\frac{\theta }{2}+1+{\tan }^2\frac{\theta }{2}}{1-{\tan }^2\frac{\theta }{2}-1-{\tan }^2\frac{\theta }{2}}=\frac{1-{\tan }^2\frac{u}{2}-e(1+{\tan }^2\frac{u}{2})+1+{\tan }^2\frac{u}{2}-e(1-{\tan }^2\frac{u}{2})}{1-{\tan }^2\frac{u}{2}-e(1+{\tan }^2\frac{u}{2})-1-{\tan }^2\frac{u}{2}+e(1-{\tan }^2\frac{u}{2})}$

$\Rightarrow \tan \frac{\theta }{2}=\pm \sqrt{\frac{1+e}{1-e}}\tan \frac{u}{2}$