高一证明题，会的请进

sin^2α+sin^2β-sin^2αsin^2β+cos^2αcos^2β
=sin^2α+sin^2β-sin^2αsin^2β+(1-sin^2α)(1-sin^2β)
=sin^2α+sin^2β-sin^2αsin^2β+1-sin^2β-sin^2α+sin^2αsin^2β
=1

多次凑成公式sin^2α+cos^2α=1就可以了。下面的步骤够详细了：
sin^2α+sin^2β-sin^2αsin^2β+cos^2αcos^2β
=（sin^2α-sin^2αsin^2β）+sin^2β+cos^2αcos^2β
=sin^2α（1-sin^2β）+sin^2β+cos^2αcos^2β
=sin^2α（1-sin^2β）+sin^2β+cos^2αcos^2β
=sin^2αcos^2β+cos^2αcos^2β +sin^2β
=cos^2β（sin^2α+cos^2α）+sin^2β
=cos^2β+sin^2β
=1

sin^2α+sin^2β-sin^2αsin^2β+cos^2αcos^2β
=（sin^2α-sin^2αsin^2β）+sin^2β+cos^2αcos^2β
=sin^2α（1-sin^2β）+sin^2β+cos^2αcos^2β
=sin^2α（1-sin^2β）+sin^2β+cos^2αcos^2β
=sin^2αcos^2β+cos^2αcos^2β +sin^2β
=cos^2β（sin^2α+cos^2α）+sin^2β
=cos^2β+sin^2β
=1
主要考查你二倍角公式的灵活运用，
其实只要会将二倍角公式运用灵活点，随便题目怎么变都不怕的，因为题目性质都一样！