求证明三角函数等式啊 1.tanα+cosα/(1+sinα)=secα 2.cotα/(1-tanα)+tanα/(1
1个回答
1.
tanα+cosα/(1+sinα)
= sinα/cosα+cosα/(1+sinα)
= [ sinα(1+sinα)+cos^2α ] / [(1+sinα)cosα ]
= [ sinα+sin^2α+cos^2α ] / [(1+sinα)cosα ]
= [ sinα+1 ] / [(1+sinα)cosα ]
= 1/cosα
= secα
2.
cotα/(1-tanα)+tanα/(1-cotα)
= cotα/(1-tanα)+tan^2α/(tanα-1)
= cotα/(1-tanα)-tan^2α/(1-tanα)
= (1/tanα-tan^2α)/(1-tanα)
= 1/tanα (1-tan^3α)/(1-tanα)
= 1/tanα (1-tanα)(1+tanα+tan^2α)/(1-tanα)
= 1/tanα (1+tanα+tan^2α)
= 1/tanα +1 + tanα
=1+tanα+cotα
相关问题
