已知α是第二象限角,且sin(π+α)=k-1/k+1,sin(5π/2+α)=3k-1/k+1,求α得正切值

由诱导公式,sin(π+α)=-sinα=(k-1)/(k+1),即sinα=(1-k)/(k+1)；
sin(5π/2+α)=cosα=(3k-1)/(k+1)
sin²α+cos²α=1；
即：(1-k)²/(k+1)²+(3k-1)²/(k+1)²=1
(1-k)²+(3k-1)²=(k+1)²
10k²-8k+2=k²+2k+1
9k²-10k+1=0
(9k-1)(k-1)
k=1/9 或 k=1
k=1/9时,sinα=(1-k)/(k+1)=4/5,cosα=(3k-1)/(k+1)=-3/5；则tanα=-4/3；且满足是第二象限角.
k=1时,sinα=(1-k)/(k+1)=0,不满足是第二象限角,舍去；
综上,α的正切值是-4/3；
如果不懂,请Hi我,

sin(π+α)=(k-1)/(k+1)
-sinα=(k-1)/(k+1)
sin(5π/2+α)=3k-1/k+1
-cosα=(3k-1)/(k+1)
tanα=sinα/cosα=(k-1)/(3k-1)

sin(π+α)=-sina=k-1/k+1
sina=-(k-1)/(k+1) >0
sin(5π/2+α)=-cosa=(3k-1)/(k+1)
cosa=-(3k-1)/(k+1) <0
sin^2a+cos^2a=1
(k-1)^2+(3k-1)^2=(k+1)^2
9k^2-10k+1=0
k=1(舍)或k=1/9
sina=4/5
cosa=-3/5
tana=-4/3

sin(π+α)=k-1/k+1,sin(5π/2+α)=3k-1/k+1
sina=-sin(π+a)=(1-k)/(k+1)
cosa=sin(π/2+a)=sin(5π/2+α)=3k-1/k+1
tana=sina/cosa=(1-k)/(3k-1)