Question

(1-sin^6θ-cos^6θ)/(1-sin^4θ-cos^4θ)

Answer 1

1-sin^6θ-cos^6θ
=1-[(sin^2θ)^3+(cos^2θ)^3]
=1-(sin^2θ+cos^2θ)(sin^4θ-sin^2θcos^2θ+cos^4θ)
=1-(sin^4θ-sin^2θcos^2θ+cos^4θ)
=1-(sin^4θ+2sin^2θcos^2θ+cos^4θ-3sin^2θcos^2θ)
=1-[(sin^2θ+cos^2θ)^2-3sin^2θcos^2θ]
=3sin^2θcos^2θ

1-sin^4θ-cos^4θ
=1-(sin^4θ+cos^4θ)
=1-[(sin^4θ+2sin^2θcos^2θ+cos^4θ-2sin^2θcos^2θ)]
=1-[(sin^2θ+cos^2θ)^2-2sin^2θcos^2θ]
=2sin^2θcos^2θ

所以：(1-sin^6θ-cos^6θ)/(1-sin^4θ-cos^4θ)
=(3sin^2θcos^2θ)/（2sin^2θcos^2θ）
=3/2。