In a triangle, if the angles `A , B ,a n dC` are in A.P. show that `2cos1/2(A-C)=(a+c)/(sqrt(a^2-a c+c^2))`

Explanation

Since angles A, B, and C are in A.P., we have `A + C = 2B`
But, `A + B + C = 180^(@) " or " 3B = 180^(@) " or " B = 60^(@)`
Now, `cos B = (1)/(2) = (a^(2) + c^(2) - b^(2))/(2ac)`
or `a^(2) + c^(2) - b^(2) = ac`
or `a^(2) - ac + c^(2) = b^(2)`
`:. (a + c)/(sqrt(a^(2) - ac + c^(2))) = (a + c)/(b)`
`= (2R(sin A + sin C))/(2R sin B)`
`= (2 sin ((A+C)/(2)) cos((A - C)/(2)))/(sin B)`
`= (2 sin 60^(@))/(sin 60^(@)) cos ((A - C)/(2))`
`= 2 cos ((A - C)/(2))`