在三角形ABC中,sinA+cosA等于二根号二,AC=2,AB=3,求tanA的值和三角形ABC的面积

(1)sinA+cosA
=√2（√2/2sinA+√2/2cosA)
=√2(sinAsinπ/4+√2/2cosA)
=√2sin(A+π/4)
=√2/2
sin(A+π/4)=1/2
且A+π/4＞π/4
因此A+π/4=5π/6,A=7π/12
tanA=tan7π/12
=tan(π/4+π/3)
=(tanπ/4+tanπ/3)/(1-tanπ/4tanπ/3)
=(1+√3)/(1-√3)
=-2-√3
(2)sinA=sin(π/4+π/3)
=sinπ/4cosπ/3+cosπ/4sinπ/3
=√2/2×1/2+√2/2×√3/2
=（√2+√6）/4
S△ABC=1/2×AB×AC×sinA
=1/2×3×2×（√2+√6）/4
=3（√2+√6）/4

sinA+cosA=√2（sinAcosπ/4+cosAsinπ/4)=√2sin(A+π/4）=√2/2
∴sin(A+π/4）=1/2
∴A=105°
∴tanA=tan105°=tan（60°+45°）= -(2+√3)

sinA+cosA等于二根号二打错了吧？
请重打一遍