How prove $A=B=C$? in $\Delta ABC$,
such

$$\sin{A}+\cos{B}+\tan{C}=\dfrac{3\sqrt{3}+1}{2}$$

prove that
$$A=B=C=\dfrac{\pi}{3}$$
My try: 
use

$$\sin{x}+\sin{y}=2\sin{\dfrac{x+y}{2}}\cos{\dfrac{x-y}{2}}$$
  then 
  \begin{align*}&\sin{A}+\cos{B}\\
&=\sin{A}+\sin{(\dfrac{\pi}{2}-B)}=2\sin{\dfrac{\pi+2(A-B)}{4}}\cos{\dfrac{A+B-\dfrac{\pi}{2}}{2}}\\
&=2\sin{\dfrac{\pi+2(A-B)}{4}}\cos{\dfrac{\pi-2C}{4}}
\end{align*}
  my idea is take $\sin{A}+\cos{B}\le f(C)$?and if only if $A=B$,BUt I can't ,Thank you someone can help me


Fix $A \in (0, \frac{\pi}{2}), A \neq \frac{\pi}{3}$. Let $f: [0, \frac{\pi}{2}) \to \mathbb R$ be
$$f(x)= \sin(A) +\tan(x)+\cos(\pi-A-x) \,.$$
As  $f$ is continuous, $f(0)=\sin(A)+\cos(\pi-A) \leq 2 < \dfrac{3\sqrt{3}+1}{2} $ and $\lim_{x \to \frac{\pi}{2}^-} f(x)= +\infty$, by the Intermediate Value Theorem there exists some $x$ so that
$$f(x)=\dfrac{3\sqrt{3}+1}{2} \,.$$
Then $A=A, B= \pi-A -x, C=x$ is a counterexample to your problem. So you cannot.
P.S. We actually prove something stronger: the angle $A$ can take any value in $(0, \frac{\pi}{2})$.
