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Question

If AB=π4, then (1+tanA)(1tanB)=

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Solution

(1+tanA)(1tanB)=1+tanAtanBtanAtanB

we know

tan(AB)=tanAtanB1+tanAtanB

1=tanAtanB1+tanAtanB

tanAtanB=1+tanAtanB

(1+tanA)(1tanB)=1+tanAtanBtanAtanB

=1+1+tanAtanBtanAtanB

=2

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