Question

已知tanα=3,则2/3sin^2α+4cos^α=

已知tanα=3,则2/3sin^2α+4cos^2α=

Answer 1

tanα=3
sinα/cosα=3
sinα=3cosα；代入：
sin²α+cos²α=1
10cos²α=1
cos²α=1/10

2/3sin²α+4cos²α
=6cos²α+4cos²α
=10cos²α
=1

Answer 2

= （2sin^2a + 2cos^2a)/(3sin^2a + 4cos^2a) 各项除以cos^2a
= ( 2 tan^2a + 2)/( 3tan^2a +4)
= ( 18 + 2)/ (27 +4)
= 20/31