sin(α-180°)+tan(315-β)/tan(β+45°)+cos(α+270°)的值为

sin(α-180°)+tan(315-β)/tan(β+45°)+cos(α+270°)的值为 sinα>tanα 化简：sin^(-α)-tan(360-α)tan(-α)-sin(180-α)cos(360-α)tan(180+α) 设cos(180°+α)=1/3 且 sinα>α 求sin(180°+α)-tanα/cos(180°-α)+tan(180°-α） 若sinαtanα 若sinα*tanα 证明三角恒等式tanαsinα/(tanα-sinα)=(tanα+sinα)/tanαsinα 证明:(tanα*sinα)/(tanα-sinα)=(tanα+sinα)/(tanα*sinα) 证明:(tanα*sinα)/(tanα-sinα)=(tanα+sinα)/(tanα*sinα) 求证(tanα·sinα)/(tanα-sinα)=(tanα+sinα)/(tanα·sinα) 求证(tanαsinα)/(tanα-sinα)=(tanα+sinα)/(tanαsinα) 求证:(tanα+tanβ)/(tanα-tanβ)=sin(α+β)/sin(α-β) .[1/cos(-α)+cos(180°+ α )]/[1/sin(540°-α)+sin(360°-α)]=tan^3α [sin(180°)-θsin(270°- θ)tan(90°- θ)]/[sin(90°+ θ)tan(270°- θ)tan(360°- θ)] 化简,cos²(-α)-tan(360°+α)/sin(-α) 化简cos^2(-α)-[tan(360°+α)/sin(-α)] 已知tanα=2,计算:[sin(90°+α)-cos(180°-α)]/[sin(90°-α)-sin(180°-α)] ①sin(α+180°)cos(-α)sin(-α-180°)和②sin³（-α）cos（2π+α）tan（-α-π）化简