已知tanα+sinα=a，tanα-sinα=b，求证: (a²-b²)²=16ab。
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左=(a^2-b^2)^2
=(a+b)^2·(a-b)^2
=[(tanα+sinα)+(tanα-sinα)]^2·[(tanα+sinα)-(tanα-sinα)]^2
=[2tanα]^2·[2sinα]^2
=16[tanα·sinα]^2
=16(tanα)^2·(sinα)^2;
右=16ab
=16(tanα+sinα)·(tanα-sinα)
=16[(tanα)^2-(sinα)^2]
=16[(sinα)^2/(cosα)^2-(sinα)^2]
=16(sinα)^2[(secα)^2-1]
=16(sinα)^2·(tanα)^2
∴左=右
=(a+b)^2·(a-b)^2
=[(tanα+sinα)+(tanα-sinα)]^2·[(tanα+sinα)-(tanα-sinα)]^2
=[2tanα]^2·[2sinα]^2
=16[tanα·sinα]^2
=16(tanα)^2·(sinα)^2;
右=16ab
=16(tanα+sinα)·(tanα-sinα)
=16[(tanα)^2-(sinα)^2]
=16[(sinα)^2/(cosα)^2-(sinα)^2]
=16(sinα)^2[(secα)^2-1]
=16(sinα)^2·(tanα)^2
∴左=右
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