# If costheta=(ecosphi-1)/(e-cosphi), prove that tan^2(theta/2)=(e+1)/(e-1)tan^2(phi/2)?

Feb 4, 2017

I started with what is given, and could prove,

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

#### Explanation:

I started with what is given, and could prove,

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

Feb 4, 2017

#### Explanation:

I think it should be $\cos \theta = \frac{e \cos \phi - 1}{e - \cos \phi}$

As $\cos \frac{\theta}{1} = \frac{e \cos \phi - 1}{e - \cos \phi}$

applying componendo & dividendo, we get

$\frac{1 + \cos \theta}{1 - \cos \theta} = \frac{e - \cos \phi + e \cos \phi - 1}{e - \cos \phi - e \cos \phi + 1}$

= $\frac{e \left(1 + \cos \phi\right) - 1 \left(1 + \cos \phi\right)}{e \left(1 - \cos \phi\right) + 1 \left(1 - \cos \phi\right)}$

= $\frac{\left(e - 1\right) \left(1 + \cos \phi\right)}{\left(e + 1\right) \left(1 - \cos \phi\right)}$

= ((e-1)(2cos^2(phi/2)))/((e+1)(2sin^2(phi/2))

= $\frac{e - 1}{e + 1} {\cot}^{2} \left(\frac{\phi}{2}\right)$

But $\frac{1 + \cos \theta}{1 - \cos \theta} = \frac{2 {\cos}^{2} \left(\frac{\theta}{2}\right)}{2 {\sin}^{2} \left(\frac{\theta}{2}\right)} = {\cot}^{2} \left(\frac{\theta}{2}\right)$

i.e. ${\cot}^{2} \left(\frac{\theta}{2}\right) = \frac{e - 1}{e + 1} {\cot}^{2} \left(\frac{\phi}{2}\right)$

Hence taking reciprocal of both sides

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