Question

证明(2-cos^2θ)(2+tan^2θ)=(1+2tan^2θ)(2-sin^2θ)

Answer 1

(2-sin^2θ)/(2-cos^2θ)
=(2(sin^2θ+cos^2θ)-sin^2θ)/(2(sin^2θ+cos^2θ)-cos^2θ)
=(sin^2θ+2cos^2θ)/(2sin^2θ+cos^2θ) 上下同时除以cos^2θ
=(2+tan^2θ)/(1+2tan^2θ)
那么就是：
(2-cos^2θ)(2+tan^2θ)=(1+2tan^2θ)(2-sin^2θ)

Answer 2

左边=(2-cos^2θ)(2+tan^2θ)
=（2-cos²θ)(2+sin²θ/cos²θ)
= (2-1+sin²θ)(2cos²θ+sin²θ)/cos²θ
= (1+sin²θ)(cos²θ+1)/cos²θ
= (cos²θ+2sin²θ)/cos²θ*(1-sin²θ+1)
=(1+2sin²θ/cos²θ)(2-sin²θ)
=(1+2tan²θ)(2-sin²θ)
=右边
∴ (2-cos^2θ)(2+tan^2θ)=(1+2tan^2θ)(2-sin^2θ)

Answer 3

左边=(1+1-cos^2θ)(1 + cos^2θ \cos^2θ + sin^2θ\cos^2θ)= (1 + sin^2θ)[1+(sin^2θ + cos^2θ)\cos^2θ)] = (1 + sin^2θ)(1+1\cos^2θ) = 1 + sin^2θ + 1\cos^2θ + sin^2θ\cos^2θ
= 1 + 2sin^2θ\cos^2θ - sin^2θ\cos^2θ + 1\cos^2θ + sin^2θ
=1 + 2tan^2θ + (1-sin^2θ)\cos^2θ + sin^2θ = 1 + 2tan^2θ + cos^2θ\cos^2θ + sin^2θ
=2+2tan^2θ + sin^2θ
右边=(1 + 2sin^2θ/cos^2θ)(1+cos^2θ) = 1 + cos^2θ + 2sin^2θ/cos^2θ + 2sin^2θ
= 1 + cos^2θ + sin^2θ + 2sin^2θ/cos^2θ + sin^2θ
= 2 + 2tan^2θ + sin^2θ
所以左边＝右边　等式成立