amath

(sin²θ ) / (2+7cos²θ) = 1/9
(1-cos^2 θ)/(2+7cos^2 θ) = 1/9
9-9cos^2 θ = 2+7cos^2 θ
16cos^2 θ = 7
cos^2 θ = 7/16
cos θ = sqrt(7)/4
tan θ = -sqrt(16-7)/sqrt(7) [Note that 3π/2< θ< 2π, tanθ <0]
tan θ = -3/sqrt(7) = -3sqrt(7)/7

(sin²θ ) / (2+7cos²θ) = 1/9
9sin²θ =2+7cos²θ
9sin²θ=2(sin²θ+cos²θ)+7cos²θ
7sin²θ=9cos²θ
sin²θ/cos²θ=9/7

你明啦

tanθ=+/-1.285714

where 3π/2&lt; θ&lt; 2π

tanθ=-1.285714

有d野打唔到出來
當見到有integer 同sin cos 時
用sin²θ +cos²θ=1 折個integer 好好用的

2007-11-23 22:01:00 補充：
咁錯機tanθ=-1.13389