Question

请大家帮我看看这道题第三问我的做法是哪里出错了？

为什么我我会解出两个tan 不知道怎么舍掉不对的那一个

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Answer 1

(1)

sinA+cosA= 1/5

(sinA+cosA)^2= 1/25

1+2sinA.cosA = 1/25

sinA.cosA = -12/25

(2)

sinA.cosA = -12/25

sin2A = -24/25

π/2<A<3π/2

=>

△ABC 是钝角三角形

(3)

sinA+cosA= 1/5

tanA +1 = (1/5)secA

(tanA+1)^2 =(1/25)(secA)^2

24(tanA)^2 + 50tanA +24 =0

12(tanA)^2 + 25tanA +12 =0

(4tanA+3)(3tanA+4)=0

tanA =-3/4(rej) or -4/3

case 1 : tanA =-3/4

sinA = 3/5 , cosA = -4/5

sinA +cosA = -1/5

case 1: 舍去

case 2 : tanA =-4/3

sinA = 4/5 , cosA = -3/5

sinA +cosA = 1/5

ie

tanA =-4/3