I have this equation \(\displaystyle{\frac{{{1}-{e}^{{2}}}}{{{1}+{e}{\cos{{\left(\theta-\theta_{{{0}}}\right)}}}}}}={1}-{e}{\cos{\eta}}\) and using the identity

London Douglas

London Douglas

Answered question

2022-04-01

I have this equation
1e21+ecos(θθ0)=1ecosη
and using the identity
cosθ=1tan2θ21+tan2θ2
this becomes
1etan(θθ02)=1+etanη2
However, I'm unsure about the steps in between.

Answer & Explanation

Boehm98wy

Boehm98wy

Beginner2022-04-02Added 18 answers

Set u=tan(θθ02) and v=tan(η2) for simplicity.
The left hand side becomes
1e21+ed1u21+u2=(1e2)(1+u2)(1+e)+(1e)u2
The right hand side becomes
1e1v21+v2=(1e)+(1+e)v21+v2
It's better to set M=1-e and P=1+e, so we get a simpler identity
MP(1+u2)P+Mu2=M+Pv21+v2
that can be rearranged to
MP+MPv2+MP(1+v2)u2=MP+P2v2+M2u2+MPu2v2
Pull all terms with u2 on the left hand side
(MP+MPv2M2MPv2)u2=P2v2MPv2
that simplifies to
M(PM)u2=P(PM)v2
Assuming PM=1+e1+e=2e0 this becomes
(1e)tan2θθ02=(1+e)tan2η2

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