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If `t a ntheta/2=sqrt((a-b)/(a+b))tanvarphi/2`, prove that `costheta=(a cosvarphi+b)/(a+b cosvarphi)`. |
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Answer» `L.H.S. = cos theta = (1-tan^2(theta/2))/(1+tan^2(theta/2))` `=(1-((a-b)/(a+b)tan^2(phi/2)))/(1+((a-b)/(a+b)tan^2(phi/2)))` `=(1-((a-b)/(a+b)sin^2(phi/2)/cos^2(phi/2)))/(1+((a-b)/(a+b)sin^2(phi/2)/cos^2(phi/2)))` `=((a+b)cos^2(phi/2)-(a-b)sin^2(phi/2))/((a+b)cos^2(phi/2)+(a-b)sin^2(phi/2))` `=(a(cos^2(phi/2) - sin^2(phi/2))+b(cos^2(phi/2) + sin^2(phi/2)))/(a(cos^2(phi/2) + sin^2(phi/2))+b(cos^2(phi/2) - sin^2(phi/2)))` `=(acosphi+b)/(a+bcosphi)` `=R.H.S.` |
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