Question #e51b8

Use the trig identity:
`tan (a + b) = (tan a + tan b)/(1 - tan a.tan b)`
Call t = (x/2) and develop the left side
`LS = tan (t + pi/4) = (tan t + tan (pi/4))/(1 - tan t.tan (pi/4))`
Trig table gives `tan (pi/4) = 1`, there for:
`LS = (1 + tan t)/(1 - tan t) = ((cos t + sin t)/(cos t))(cos t/(cos t - sin t)) =`
`LS = (cos t + sin t)/(cos t - sin t)`
Multiply both numerator and denominator by (cos t + sin t), we get:
`LS = (cos t + sin t)^2/(cos^2 t - sin^2 t)`
Reminder:
`(cos t + sin t)^2 = 1 + 2cos t.sin t = 1 + sin 2t`
`cos^2 t - sin^2 t = cos 2t`.
`LS = (1 + sin 2t)/(cos 2t) = 1/(cos 2t) + (sin 2t)/(cos 2t)`
`LS = sec 2t + tan 2t`
Replace t by `(x/2)`, we get
`tan (x/2 + pi/4) = sec x + tan x`

`RHS=secx+tanx`

`=1/cosx+sinx/cosx`

`=(1+sinx)/cosx`

`=(cos^2(x/2)+sin^2(x/2)+2sin(x/2)cos(x/2))/(cos^2(x/2)-sin^2(x/2))`

`=(cos(x/2)+sin(x/2))^2/((cos(x/2)+sin(x/2))(cos(x/2)-sin(x/2)))`

`=(cos(x/2)+sin(x/2))/(cos(x/2)-sin(x/2))`

`=(cos(x/2)/cos(x/2)+sin(x/2)/cos(x/2))/(cos(x/2)/cos(x/2)-sin(x/2)/cos(x/2))`

`=(1+tan(x/2))/(1-tan(x/2))`

`=(tan(pi/4)+tan(x/2))/(1-tan(x/2)tan(pi/4))`

`=tan(x/2+pi/4)=LHS`

`tan (x/2 +pi/4)= (tan pi/4 + tan x/2)/(1- tan x/2 tan pi/4)`

=`(1+ tan x/2)/(1-tan x/2)` = `(cos x/2 +sinx/2)/(cos x/2- sin x/2)`

Now multiply the numerator and denominator by `(cos x/2 + sin x/2)`

= `(1+ 2sin x/2 cos x/2)/(cos^2 x/2 - sin^2 x/2)` = `(1+sin x)/cos x`= sec x + tanx