Question

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

Answer 1

See the explanation below

Explanation:

We need

`costheta=2cos^2(theta/2)-1`

`cos(theta/2)=+-sqrt((1+costheta)/2)`

`costheta=1-2sin^2theta`

`sin(theta/2)=+-sqrt((1-costheta)/2)`

`tan(theta/2)=+-sqrt((1-costheta)/(1+costheta))`

`costheta=(cosphi-e)/(1-ecosphi)`

Therefore,

`tan(theta/2)=+-sqrt((1-((cosphi-e)/(1-ecosphi)))/(1+((cosphi-e)/(1-ecosphi))))`

`=+-sqrt((1-ecosphi-cosphi+e)/(1-ecosphi+cosphi-e))`

`=+-sqrt(((1+e)(1-cosphi))/((1-e)(1+cosphi)))`

`=+-sqrt((1+e)/(1-e))*sqrt((1-cosphi)/(1+cosphi))`

But,

`tan(phi/2)=+-sqrt((1-cosphi)/(1+cosphi))`

Therefore,

`tan(theta/2)=+-sqrt((1+e)/(1-e))tan(phi/2)`

Answer 2

Please see below.

Explanation:

Recall the identity `tan(theta/2)=+-sqrt((1-costheta)/(1+costheta))`. Note that this emerges from another identity `costheta=1-2sin^2(theta/2)=2cos^2(theta/2)-1`

Now as `costheta=costheta/1=(cosphi-e)/(1-ecosphi)`, using componendo dividendo, we get

`(costheta+1)/(costheta-1)=(cosphi-e+1-ecosphi)/(cosphi-e-1+ecosphi)`

`=(1+cosphi-e(1+cosphi))/(e(cosphi-1)-1+cosphi)`

`((1+cosphi)(1-e))/((1+e)(cosphi-1))`

Hence `(1-costheta)/(1+costheta)=((1+e)(1-cosphi))/((1-e)(1+cosphi))`

and `tan(theta/2)=+-sqrt(((1+e)(1-cosphi))/((1-e)(1+cosphi)))`

`= +-sqrt((1+e)/(1-e))sqrt((1-cosphi)/(1+cosphi))`

`= +-sqrt((1+e)/(1-e))tan(phi/2)`

Note: Application of componendo and dividendo means if `a/b=c/d`, then `(a+b)/(a-b)=(c+d)/(c-d)`