If sin (a + b) = 1, then Tan (2a + b) + tanb =?

sin(A+B)=1
A+B=2kπ+π/2
2A+2B=4kπ+π
tan(2A+2B)=tan(4kπ+π)=0
tan[(2A+B)+B]=0
So [Tan (2a + b) + tanb] / [1-tan (2a + b) * tanb] = 0
So tan (2a + b) + tanb = 0

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