证明：
左边=4+2tan^2α-2cos^2α - cos^2α tan^2α
=2+2tan^2α+2-2cos^2α-sin^2α
=2+2tan^2α+2sin^2α-sin^2α
=2 + 2tan^2α + sin^2α
右边=（1+2tan^2α)(1+cos^2α)
=1+cos^2α+2tan^2α+2tan^2αcos^2α
=1+cos^2α+2tan^2α+2sin^2α
=1+1-sin^2α+2tan^2α+2sin^2α
=2+2tan^2α+sin^2α
左边等于右边原式成立.