\sin A+\tan A = \dfrac{2\tan \dfrac{A}{2} }{1+\tan^2 \dfrac{A}{2} }+ \dfrac{2\tan \dfrac{A}{2} }{1-\tan^2 \dfrac{A}{2} }

= 2\tan \dfrac{A}{2}\Bigg( \dfrac{1}{1+\tan^2 \dfrac{A}{2}}+\dfrac{1}{1-\tan^2 \dfrac{A}{2}} \Bigg)

Since A is an acute angle,

0< A < \dfrac{\pi}{2}

0< \dfrac{A}{2} < \dfrac{\pi}{4}

0< \tan \dfrac{A}{2} < 1

Therefore,

\dfrac{\tan \dfrac{A}{2}}{1-\tan^4 \dfrac{A}{2}} >\tan \dfrac{A}{2}

Since f(x)=\tan x is always above f(x) =x when x >0

We get, \tan \dfrac{A}{2} > \dfrac{A}{2}

From (1),

Similarly,

Addition of (2), (3) and (4)

\sin A+ \sin B + \sin C+\tan A+\tan B+\tan C>2(A+B+C) = 2\pi