Question

To prove; (1-e) tan^2B/2=(1+e) tan^2A/2=(cosA-e)/(1-e cosA)​

Answer 1

Answer:Here, tan[

θ

2

 

]=

√

1-e

1+e

 

tan(

ϕ

2

 

)

we know, tan(

θ

2

 

)=±

√

1-cosθ

1+cosθ

 

So, our equation becomes,

⇒±

√

1-cosθ

1+cosθ

 

=

√

1-e

1+e

 

(±

√

1-cosθ

1+cosθ

 

)

⇒(

1-cosθ

1+cosθ

 

)=(

1-e

1+e

 

)(

1-cosθ

1+cosθ

 

)

⇒(1+e)(1-cosθ)(1+cosϕ)=(1-e)(1+cosθ)(1-cosϕ)

⇒(1+e)(1+cosϕ-cosθ-cosθcosϕ)=(1-e)(1-cosϕ+cosθ-cosθcosϕ)

⇒1+cosϕ-cosθ-cosϕcosθ+e+ecosϕ-ecosθ-ecosθcosϕ=1-cosϕ+cosθ-cosϕcosθ-e+ecosϕ-ecosθ+ecosθcosϕ

⇒2cosϕ+2e=2cosθ+2ecosθcosϕ

⇒cosϕ+e=cosθ+ecosθcosϕ

⇒cosϕ(1-ecosθ))=cosθ-e

⇒cosϕ=

cosθ-e

1-ecosθ

 

Step-by-step explanation: