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)#