化简 [ sin(270°+B) × sin(90°-B) × cot(B-270°) ] / [ sin(-B) × cos(270°-B) × cot(180°-B) ]

题目

化简 [ sin(270°+B) × sin(90°-B) × cot(B-270°) ] / [ sin(-B) × cos(270°-B) × cot(180°-B) ]
要完整,

答案

[sin(270°+B) × sin(90°-B) × cot(B-270°) ] / [ sin(-B) × cos(270°-B) × cot(180°-B)]
=[-cosBcosB(-tanB)]/[(-sinB)(-sinB)(-cotB)]
=-[cosBcosBtanB]/[sinBsinBcotB]
=-cotBcotBtanB/cotB
=-1