(x+sin∧2x+tanx)/(sinx+x∧2)在x→0时的极限

∫sin^2x(1+tanx)dx (tanx-sinx)/x^2sin^2x 极限x--->0 求化简(sinx+tanx)/cos^2x+sin^2x+cosx f(x)=(1+1/tanx)sin(x)^2化简 已知函数f(x)=(1+1 anx)sin^2x+m sin(x+π/4)sin(x-π/4) 30《x《60,求y=(tanx-sin^2x)/(tanx+sin^2x)的最大值最小值. 提问数学难题求证：sin^2x*tanx+cos^2x/tanx+2sinx*cosx=tanx+1/tanx 若tanx=2,则2sin平方x-3sin（π-x）sin（π/2-x）= tanx/2=1/2,求sin(x+π/6) 化简(sin2x+2sin^2 x)/1-tanx (2sin^2X+sin2X)/(1+tanX)怎么化简? 如果tanx=2,那么sin^2(x)+sinx*cosx tanx/2=1/2求sin(x+π/6) 求极限lim.[( tanx-sinx) /(sin^2 2x)] 已知tanx 2 则sin^2x-sinxcosx 已知tanx=2,则sin^2x+1=? 已知tanx=2,则sin^2x+1=? 化简(sin^2x+2sinxcosx)/1+tanx