化简:(1)−sin(π+α)+sin(−α)−tan(2π+α)tan(α−π)+cos(−α)+cos(π−α)
证明:1+sinα−cosα1+sinα+cosα=tanα2
若tanα=2,求2sinα+cosαsinα−cosα
已知tanα=2,求sinα−3cosαsinα+cosα
化简:[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(
设tan(5π+α)=m,则sin(α−3π)+cos(π−α)sin(−α)−cos(π+α)
已知tanα=3,求2cos(π−α)−3sin(π+α)4cos(−α)+sin(2π−α)
若sinα+cosα2sinα−cosα=2,则tanα=( )
若tanα=2,则2sinα−cosαsinα+cosα的值为( )
计算:(Ⅰ)若tanα=-2,求1+2sin(π−α)sin(3π2+α)cos
求证:[sin^2(π+α)*cos(π/2-α)+tan(2π-α)*cos(-α)]/-sin^2(-α)+tan(
化简:tanα*(cosα-sinα)+[sinα(sinα+tanα)/1+cosα]
化简:tanα(cosα-sinα)+sinα(sinα+tanα)/1+cosα.