Question

一道数学题，求解。（帮帮忙！谢谢了！）

已知tanα=2，求（sin2α-cos2α）/（cos^2）α的值。

Answer 1

致学

Answer 2

原式=（2sin acos a-(cos^2) a+(sin^2) a）/(cos^2）α
=2tan a-1+(tan^2) a
=2*2-1+2^2
=7

楼上最后结果算错了

Answer 3

原式=（2sinacosa-(cos^2)a+(sin^2)a）/(cos^2）α
=2tana-1+(tan^2)a=4-1+4=-1

Answer 4

（sin2α-cos2α）/（cos^2）α=(2sinacosa-(cos^2-sin^2))/（cos^2）α=2tana-(1-tan^2a)=7

Answer 5

=[2sinαcosα-(cos^2α-sin^2α)]/cos^2α
=tan^2α+2tanα-1
=(1+tanα)^2-2
=(1+2)^2-2
=9-2
=7

Answer 6

tanα=2，求（sin2α-cos2α）/（cos^2）α
（sin2α-cos2α）/（cos^2）α=（2sinαcosα-2（cos^2）a+1）/（cos^2）α=（2tanα-2+1/（cos^2）a）=（2tanα-2+1+tan^2α）=4-2+1+4=7