यदि `"tan"(phi)/(2)=sqrt((1-e)/(1+e))"tan"(theta)/(2)` हो, तो सिद्ध कीजिए कि - `cos phi = (cos theta + e)/(1+e cos theta)`.

Explanation

चूँकि हम जानते हैं, कि
`cos theta = (1-"tan"^(2)(theta)/(2))/(1+"tan"^(2)(theta)/(2))rArr cos phi = (1-"tan"^(2)(phi)/(2))/(1+"tan"^(2)(phi)/(2))`
`=(1-((1-e)/(1+e))"tan"^(2)(theta)/(2))/(1+((1-e)/(1+e))"tan"^(2)(theta)/(2)) " " (because "tan"(phi)/(2)=sqrt((1-e)/(1+e))"tan"(theta)/(2))`
`((1+e)-(1-e)"tan"^(2)(theta)/(2))/((1+e)+(1-e)"tan"^(2)(theta)/(2))`
`=((1-"tan"^(2)(theta)/(2))+e(1+"tan"^(2)(theta)/(2)))/((1+"tan"^(2)(theta)/(2))+e(1-"tan"^(2)(theta)/(2)))`
`((1+"tan"^(2)(theta)/(2))[(1-"tan"^(2)(theta)/(2))/(1+"tan"^(2)(theta)/(2))+e])/((1+"tan"^(2)(theta)/(2))[1+e((1-"tan"^(2)(theta)/(2)))/((1+"tan"^(2)(theta)/(2)))])`
`=(cos theta + e)/(1+ e cos theta)= R.H.S. " "` इति सिद्धम्