解释cot(πx/2)=tan(π/2-πx/2)=tan[(1-x)π/2]～(1-x)π/2

设函数f(x)=[cot(-x-π)sin(2π+x)]/[cos(-x)tan(3π-x)].(1)若f(α)=（根3 1)tan(x/2+π/4)+tan(x/2-π/4)=2tanx 请问sin(x+π/2)=?cos(x+π/2)=?tan(x+π/2)=?cot(x+π/2)=?sec(x+π/2) f(a)=[sin(π-x)cos(2π-x)tan(-x+3π/2)]/cot(-x -π)sin(-π--x) 化简 设y=(tan^2 x-csc^2 x )/(tan^2 x+cot^2 x -1) (a)证明y=1- 2/(tan^ tanx+tan(π/2 -x)=? 怎么证明tan^2x+cot^2x=2(3+cos4x)/1-cos4x 化简cos(π/2-x)cos(π/2+x)cot(π-x)/sin(3π/2+x)cos(3π+x)tan(π+x) 函数f(x)=tan(x^2-π/2)是( ) 证明sec x+tanx=tan(π/4 +x/2) tan（X/2+π/4)+tan（x/2-π/4）=2tanx? tan（ x/2+π/4）+tan（x/2-π/4 ）=2tanx