Question

Show that `2tan^-1 {tan  alpha/2 * tan(pi/4 - beta/2)} = tan^-1  (sin alpha cos beta)/(cosalpha + sinbeta)`

Answer 1

L.H.S. = `tan^-1  (2tan  alpha/2 * tan (pi/4 - beta/2))/(1 - tan^2  alpha/2 tan^2 (pi/4 - beta/2))`  ......`("since"  2 tan^-1x = tan^-1  (2x)/(1 - x^2))`

= `tan^-1  (2tan  alpha/2  (1 - tan  beta/2)/(1 + tan  beta/2))/(1 - tan^2  alpha/2  ((1 - tan  beta/2)/(1 + tan  beta/2))^2)`

= `tan^-1  (2tan  alpha/2 (1 - tan^2  beta/2))/((1 + tan  beta/2)^2 - tan^2  alpha/2 (1 - tan  beta/2)^2)`

= `tan^-1  (2tan  alpha/2 (1 - tan^2  beta/2))/((1 + tan^2  beta/2)(1 - tan^2  alpha/2) + 2   beta/2 (1 + tan^2  alpha/2))`

= `tan^-1  ((2tan  alpha/2)/(1 + tan^2  alpha/2) - (1 - tan^2  beta/2)/(1 + tan^2  beta/2))/((1 - tan^2  alpha/2)/(1 + tan^2  alpha/2) + (2tan  beta/2)/(1 + tan^ beta/2))`

= `tan^-1  ((sin alpha cos beta)/(cos alpha + sin beta))`

= R.H.S.