若α,β为锐角,且α+β=45°求证(1+tanα)(1+tanβ)=2(2)求log2（1+tan1°)+log2(1+tan2°)+···log2(1+45°)=?

Question

若α,β为锐角,且α+β=45° 求证(1+tanα)(1+tanβ)=2
(2)求log2（1+tan1°)+log2(1+tan2°)+···log2(1+45°)=?

Answer 1

tan(a+b)=1
tana+tanb=1-tana*tanb
tana+tanb+tana*tanb=1
(1+tanα)(1+tanβ)
=1+tana+tanb+tana*tanb
=2
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由 1=tan45°=(tan1°+tan44°)/(1-tan1°tan44°),
得 tan1°+tan44°=1-tan1°tan44°,即tan1°+tan44°+tan1°tan44°=1,
则(1+tan1°)(1+tan44°)=1+tan1°+tan44°+tan1°tan44°=2；
同理(1+tan2°)(1+tan43°)=2,.,(1+tan22°)(1+tan23°)=2.
又1+tan45°=2,
则 (1+tan1°)(1+tan2°).(1+tan44°)(1+tan45°)
=2^23.
log2（1+tan1°)+log2(1+tan2°)+···log2(1+45°)=23