Calculate (1+tan1°)×(1+tan2°)×(1+tan3°)×...×(1+tan44°)

The answer is 4194304.

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First If A+B=45° then (1+tanA)(1+tanB)=2

Prove: we all know that

Tan(A+B)=(tanA+tanB)÷(1-tanAtanB) So,tanA+tanB=tan(A+B)-tan(A+B)tanAtanB

(1+tanA)(1+tanB) =1+tanAtanB+tanA+tanB =1+tanAtanB+tan(A+B)-tan(A+B) tanAtanB

Because A+B=45° so tan (A+B)=1 =1+tan(A+B)

Because A+B=45° so tan (A+B)=1 =1+1 =2

So,with the conclusion above,then rearrange the question into a(1+tanA)(1+tanB)form [A+B=45°]

So (1+tan1°)(1+tan44°)(1+tan2°)(1+tan43°)×... ×(1+tan22°)(1+tan23°)=2^22=4194304