求证：(1/sinα-sin(180°+α))/(1/cos(540°-α)+cos(360°-α))=1/(tanα)

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求证：(1/sinα-sin(180°+α))/(1/cos(540°-α)+cos(360°-α))=1/(tanα)^3

是否是：(-1/sinα-sin(180°+α))/(1/cos(540°-α)+cos(360°-α))=1/(tanα)^3
证明:
左边=[-1/sina-(-sina)]/[1/(-cosa)+cosa]
=[(-1+sin^2a)/sina]/[(cos^2-1)/cosa]
=-cos^2a/sina*cosa/(-sin^2a)
=cos^3a/sin^3a
=1/(tana)^3
=右边.

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