# Verify if the identity holds if x is in [0,pi/2]. (1+sinx)/(1+cosx)=(tan(x/2))'+tan(x/2)

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### 1 Answer

We'll manage the right side of the expression and we'll differentiate tan(x/2).

[tan(x/2)]' = 1/2*[cos(x/2)]^2

We'll re-write the expression:

(1+sinx)/(1+cosx)=1/2*[cos(x/2)]^2 + tan(x/2)

We'll replace the term tan (x/2) by the fraction sin (x/2)/cos (x/2).

(1+sinx)/(1+cosx)=1/2*[cos(x/2)]^2 + sin (x/2)/cos (x/2)

We'll multiply the term sin (x/2)/cos (x/2) by 2*cos (x/2):

(1+sinx)/(1+cosx)=[1 + 2*cos (x/2)*sin(x/2)]/2*[cos(x/2)]^2

We'll recognize the formula of double angle:

2*cos (x/2)*sin(x/2) = sin 2*(x/2) = sin x

(1+sinx)/(1+cosx)=(1+sinx)/2*[cos(x/2)]^2

We'll replace the denominator from the right side by the formula of half angle:

(1+sinx)/(1+cosx)=(1+sinx)/(1+cosx)

**We notice that managing RHS, we'll get LHS, therefore the identity is verified, if x belongs to the range [0 , pi/2].**