Question

若sinα-3cosα=√5,则tan2α=？

Answer 1

sinα=√5+3cosα
两边平方
sin²α=1-cos²α=5+6√5cosα+9cos²α
5cos²α+3√5cosα+2=0
cosα=[-3√5±√(45-40)]/10
cosα=-2√5/5或-√5/5

cosα=-2√5/5时
sinα=√5+3cosα=-√5/5
tanα=sinα/cosα=1/2
tan2α=2tanα/(1-tan²α)=4/3

cosα=-√5/5时
sinα=√5+3cosα=2√5/5
tanα=sinα/cosα=-2
tan2α=2tanα/(1-tan²α)=4/3

综上
tan2α=4/3

Answer 2

sinα-3cosα=√5
tanα - 3=√5secα
(tanα - 3)^2=5(secα)^2
(tanα - 3)^2=5[(tanα)^2 +1]
4(tanα)^2 +6tanα -4 =0
2(tanα)^2 +3tanα -2 =0
(2tanα -1)(tanα+2)=0
tanα =1/2 or -2

when tanα =1/2
tan2α
=2tanα/[1- (tanα)^2 ]
=1/(1-1/4)
=4/3

when tanα =-2
tan2α
=2tanα/[1- (tanα)^2 ]
=-4/(1-4)
=4/3

ie
tan2α =4/3