If `t a ntheta/2=sqrt((a-b)/(a+b))tanvarphi/2` , prove that `costheta=(a cosvarphi+b)/(a+b cosvarphi)` .

Explanation

Given,
`tan"" (theta)/(2)=sqrt((a-b)/(a+b))tan ""(varphi)/(2)`
Now, `cos theta=(1-tan^(2)(theta)/(2))/(1+tan^(2)""(theta)/(2))=(1-(a-b)/(a+b)tan^(2)""(varphi)/(2))/(1+(a-b)/(a+b)tan^(2)""(varphi)/(2))`
`1-(a-b)/(a+b)(sin^(2)""(varphi)/(2))/(cos^(2)""(varphi)/(2))`
`=1+(a-b)/(a+b)(sin^(2)""(varphi)/(2))/(1+(a-b)/(a+b)(sin^(2)""(varphi)/(2))/(cos^(2)""(varphi)/(2))`
`=((a+b)cos^(2)""(varphi)/(2)-(a-b)sin^(2)""(varphi)/(2))/((a+b)cos^(2)""(varphi)/(2)+(a-b)sin^(2)""(varphi)/(2))`
`=(a(cos^(2)""(varphi)/(2)-sin^(2)""(varphi)/(2))+b(cos^(2)""(varphi)/(2)+sin^(2)""(varphi)/(2)))/(a(cos^(2)""(varphi)/(2)+sin^(2)""(varphi)/(2))+b(cos^(2)""(varphi)/(2)-sin^(2)""(varphi)/(2)))`
`=(a cos phi+b)/(a+b cos phi)`