<s>[INST] 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

 [/INST] The plot of
$$\sin A + \cos B + \tan \left(\pi - A - B \right)$$
indicates that the expression attains any particular value infinitely often. Consequently, while $A = B = \pi - A - B = \pi/3$ produces the value $\frac{1}{2}\left(3\sqrt{3}+1\right)$, this is far from the only solution to your equation.

</s>