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(sinx+cosx)(tan^2x+1/tanx)=1/cosx+1/sinx
人气:482 ℃ 时间:2019-10-19 18:23:12
解答
(sinx+cosx)(tan^2x+1/tanx)-[1/cosx+1/sinx]
=(sinx+cosx ) (tan^2x+1/tanx)-[(sinx+cosx)/sinxcosx]
=(sinx+cosx)[tanx+(1/tanx)-(1/sinxcosx)]
=(sinx+cosx)[(sinx/cosx)+(cosx/sinx)-(1/sinxcosx)]
=(sinx+cosx)[(1/sinxcosx)-(1/sinxcosx)]
=0
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