Question

已知tanθ=3,则sin^2θ+sinθcosθ-2cos^2θ=

Answer 1

解：
sin²θ+sinθcosθ-2cos²θ
= cos²θ(tan²θ +tanθ -2)
= (tan²θ +tanθ -2)/sec²θ
=(tan²θ +tanθ -2)/(1+tan²θ)
=(3*3 + 3 -2)/(1+3*3) =1

Answer 2

sin^2θ+sinθcosθ-2cos^2θ
=1/cos^2θ (sin^2θ/cos^2θ+sinθcosθ/cos^2θ-2cos^2θ/cos^2θ)
=1/cos^2θ (tan^2θ+tanθ-2)
= (tan^2θ+1) (3^2+3-2)
=(3^2+1)*(9+1)
=100

Answer 3

tanθ=3
sin^2θ+sinθcosθ-2cos^2θ
= 1/cos^2θ * (tan^2θ+tanθ-2）
= (tan^2θ+1) * (tan^2θ+tanθ-2）
= (3^2+1) * （3^2+3-2）
= 10 * 10
= 100

Answer 4

sin^2θ+sinθcosθ-2cos^2θ=(sin^2θ/cos^2θ+sinθ/cosθ-2)cos^2θ
=(tan^2θ+tanθ-2)/sec^2θ
=(tan^2θ+tanθ-2)/(tan^2θ+1)
=(3^2+3-2)/(3^2+1)
=1