2.3　两角和与差的正切函数

1.已知α∈("-" π/2 "," 0),sin α=-4/5,则tan(α+π/4)=0(　)

A.-7 B.-1/7 C.1/7 D.7

解析∵α∈("-" π/2 "," 0),∴cos α=3/5,

　　∴tan α=-4/3.

　　∴tan(α+π/4)=(tanα+1)/(1"-" tanα)=-1/7.

答案B

2.已知tan(α+β)=2/5,tan(β"-" π/4)=1/4,那么tan(α+π/4)=(　　)

A.13/18 B.13/22 C.3/22 D.-3/22

解析因为α+π/4=(α+β)-(β"-" π/4),

　　所以tan(α+π/4)=tan["(" α+β")-" (β"-" π/4)]

　　=(tan"(" α+β")-" tan(β"-" π/4))/(1+tan"(" α+β")" tan(β"-" π/4) )=(2/5 "-" 1/4)/(1+2/5×1/4)=3/22,

　　故选C.

答案C

3.若A=15°,B=30°,则(1+tan A)(1+tan B)的值为0(　　)

A.1 B.2 C.-1 D.-2

解析由结论A+B=45°,则(1+tan A)(1+tan B)=2.

答案B

4.若tan α=lg(10a),tan β=lg1/a,且α+β=π/4,则实数a的值为(　　)

A.1 B.1/10 C.1或1/10 D.1或10

解析tan α+tan β=lg(10a)+lg1/a=lg 10=1.

　　∵α+β=π/4,

　　∴tanπ/4=tan(α+β)=(tanα+tanβ)/(1"-" tanαtanβ)=1/(1"-" tanαtanβ)=1,∴tan αtan β=0,

　　则有tan α=lg(10a)=0或tan β=lg1/a=0,

　　∴10a=1或1/a=1,即a=1/10或1,故选C.

答案C

5.若锐角α,β使α+2β=2π/3,tanα/2tan β=1/3同时成立,则α+β的值为(　　)

A.5π/12 B.π/2 C.7π/12 D.5π/6

解析∵α+2β=2π/3,∴α/2+β=π/3,∴tan(α/2+β)=(tan α/2+tanβ)/(1"-" tan α/2 tanβ)=(tan α/2+tanβ)/(1"-" 1/3)=√3,即tanα/2+tan β=(2√3)/3,∴tanα/2,tan β是x2-(2√3)/3x+1/3=0的两个根,解得tanα/2=tan β=√3/3.

　　又α,β均为锐角,

　　∴α/2=β=π/6,故α+β=π/2.

答案B

6.(2018全国Ⅱ高考)已知tan(α"-" 5π/4)=1/5,则tan α=　　　　　　.

解析∵tan(α"-" 5/4 π)=(tanα"-" tan 5/4 π)/(1+tanαtan 5/4 π)=(tanα"-" 1)/(1+tanα)=1/5,

∴5tan α-5=1+tan α.∴tan α=3/2.