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].
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