`2tan^-1(sqrt((a-b)/(a+b) )tan (theta/2))`
`= cos^-1((acos theta+b)/(a + bcos theta))`
as we know, `2tan^-1 x = 2 tan^-1 x = cos^-1((1-x^2)/(1+x^2))`
so,LHS=`2tan^-1( sqrt((a-b)/(a+b)) tan(theta/2)) - cos^-1((1- ((a-b)/(a+b)) tan (theta/2))/(1 + ((a-b)/(a+b))tan^2 (theta/2)))`
` = cos^-1(( 1 - ((a-b)(sin^2(theta/2)))/((a+b)(cos^2(theta/2))))/(1- ((a-b)(sin^2(theta/2)))/((a+b)(cos^2 theta/2))))`
`= cos^-1(( ((a+b)cos^2 (theta/2) - (a-b)sin^2(theta/2))/((a+b)cos^2 (theta/2)))/(((a+b)cos^2(theta/2) - (a-b)sin^2(theta/2))/((a+b)cos^2(theta/2))))`
`=> cos^-1((a(cos^2 (theta/2) - sin^2( theta/2)) + b(cos^2(theta/2) + sin^2(theta/2)))/((a(cos^2(theta/2)) + sin^-1(theta/2))+ b(cos^2 (theta/2) - sin^2 (theta/2))))`
`= cos^-1((acos theta + b)/(a+bcos theta)) =`RHS
hence proved
