1)tan(x/2+π/4)+tan(x/2-π/4)=2tanx(2)(1-2sinαcosα/cos²α-sin²α)=(1-tanα)/(1+tanα)

问题描述:

1)tan(x/2+π/4)+tan(x/2-π/4)=2tanx
(2)(1-2sinαcosα/cos²α-sin²α)=(1-tanα)/(1+tanα)

1.
左=tan(x/2+π/4)+tan(x/2-π/4)
=tan[(x/2+π/4)+(x/2-π/4)][1-tan(x/2-π/4)tan(x/2+π/4)]
=tanx[1-(-1)]=2tanx=右
2.
左=1-2sinαcosα/cos²α-sin²α
=(cosα-sinα)²/(cosα-sinα)(cosα+sinα)
=(cosα-sinα)/(cosα+sinα)=(1-tanα)/(1+tanα)=右