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

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Answer 1 · 2017-02-04T15:57:47

I started with what is given, and could prove,

$tan^2(theta/2)={1-e tan^2(phi/2)}/{tan^2(phi/2)+e}$

I started with what is given, and could prove,

$tan^2(theta/2)={1-e tan^2(phi/2)}/{tan^2(phi/2)+e}$

Answer 2 · 2017-02-04T18:12:43

Please see below.

I think it should be $costheta=(ecosphi-1)/(e-cosphi)$

As $costheta/1=(ecosphi-1)/(e-cosphi)$

applying componendo & dividendo, we get

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

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

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

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

= $(e-1)/(e+1)cot^2(phi/2)$

But $(1+costheta)/(1-costheta)=(2cos^2(theta/2))/(2sin^2(theta/2))=cot^2(theta/2)$

i.e. $cot^2(theta/2)=(e-1)/(e+1)cot^2(phi/2)$

Hence taking reciprocal of both sides

$tan^2(theta/2)=(e+1)/(e-1)tan^2(phi/2)$

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