1.若α+β=45°,求证（tanα+1)(tanβ+1）=2 2.若（tanα+1）（tanβ+1）=2,求α+β的值

1.若α+β=45°,求证（tanα+1)(tanβ+1）=2 2.若（tanα+1）（tanβ+1）=2,求α+β的值
1.若α+β=45°,求证（tanα+1)(tanβ+1）=2 2.若（tanα+1）（tanβ+1）=2,求α+β的值

1.若α+β=45°,求证（tanα+1)(tanβ+1）=2 2.若（tanα+1）（tanβ+1）=2,求α+β的值
(Tana+1)(tanb+1)=tanatanb+tana+tanb+1=tanatanb+tan(a+b)(1-tanatanb)+1(a+b=45,tan(a+b)=1)=2.(tana+1)(tanb+1)=2_tanatanb+tana+tanb+1=2._tanatanb+tan(a+b)(1-tanatanb)=1.故tan(a+b)＝1.a+b=45.