Tan x/2 =tan u/2 then show that u=log tan(pi/4+x/2)

Question

Is from Mumbai University  paper 

Answer 1

tanx/2 = tanu/2 is ambiguous

it could mean tan(x/2) = tan(u/2), then u=x +/-pi

as the period for tangent x is pi

or it could mean (tanx)/2 = (tanu)/2,

then u=the same value in terms of x, u= x +/-pi

tan45)/2 = 1/2 = tan45)/2 = 1/2 = tan135)/2 = 1/2

or

tan(pi/4)/2 = 1/2 = tan(pi/4+pi)/2 = 1/2

meanwhile

u = logtan(pi/4+x/2) would have to mean

u =logtan(pi/4+u/2)

and

maybe replace log with ln

then

u = lntan(pi/4 +u/2)

then

e^u = tan(pi/4 +u/2)

tan(a+b) = (tana+tanb)/(1-tanatanb)

tan(pi/4+u/2) = (tan(pi/4) + tan(u/2))/(1-tan(pi/4)tan(u/2))

= (1+tan(u/2))/(1-tan(u/2)

= (1+tan(u/2)(1+tan(u/2))/(1-tan^2(u/2))

e^u= (1 +2tan(u/2) + tan^2(u/2))/(1-tan^2(u/2))

e^u(1-tan^2(u/2) = 1+ 2tan(u/2) + tan^2(u/2)

let z=u/2

e^2z(1-tan^2(z)) = 1+2tanz + tan^2(z)

e^2z - e^2z(tan^2(z) = 1+tanz +tan^2(z)

graph it and see where z crosses the x axis

which is only at the origin: z=0, u/2= 0, u=0=x

solution is u=x=0