Question

若tanα+(1/tanα)=3，则sinαcosα=_____，tan^2α+(1/tan^2α)=________

若tanα+(1/tanα)=3，则sinαcosα=_____，tan^2α+(1/tan^2α)=________

Answer 1

由tanα+(1/tanα)=sinα/cosα+cosα/sinα=(sin²α+cos²α)/sinαcosα=3
于是有：1/sinαcosα=3
因此，sinαcosα=1/3

(tanα+(1/tanα))²=tan²α+(1/tan²α)+2=9
故：tan^2α+(1/tan^2α)=9-2=7

Answer 2

tanα+(1/tanα)=sinα/cosα+cosα/sinα=[(sinα)^2+(cosα)^2]/(sinαcosα)
=1/(sinαcosα)=3,
∴sinαcosα=1/3。
tan^2α+(1/tan^2α）=
=[（tanα+(1/tanα）]^2-2=9-2=7.

Answer 3

tanα+(1/tanα)=(tan²α +1)/tanα= 3
sinαcosα=(sinαcosα)/(sin²α +cos²α) =tanα/(tan²α +1) =1/3
tan^2α+(1/tan^2α)=[tanα+(1/tanα)]²-2=7

Answer 4

tanα+(1/tanα)=3 , sina/cosa+cosa/sina=(sin^2a+cos^2a)/sinacosa=3 , sinacosa=1/3
[tanα+(1/tanα)]^2=9=tan^2α+2+(1/tan^2a)
tan^2α+(1/tan^2α)=7

Answer 5

⑴sinαcosα=_____
∵tanα=sinα/cosα
tanα+(1/tanα)=3
∴sinα/cosα+cosα/sinα=3 ①
又∵sinα^2+cosα^2=1
①式两端同时乘以sinαcosα得
sin^2α+cos^2α=3sinαcosα=1
∴sinαcosα=1/3

⑵tan^2α+(1/tan^2α)=________
[tanα+(1/tanα)]^2=tan^2α+(1/tan^2α)+2=9
∴tan^2α+(1/tan^2α)=9-2=7

看题按步骤写出 忘采纳 谢谢^_^

Answer 6

(1)3
将已知的式子用sin和cos展开，可以得到
(2)7
将已知的式子两端平方