若tan(x-π)=2,求2sin^2x+1/cos^2x-sin^2x的值若tan(x-π)=2,求2sin^2x+1/cos^2x-sin^2x的值

已知tan=2,求(cos x+sin x)/(cos x-sin x)+sin^2x 若tan(x-π)=2,求2sin^2x+1/cos^2x-sin^2x的值若tan(x-π)=2,求2sin^2x+1/cos^2x-sin^2x的值 求极限：sin x^tan x(x→π/2) tan(x)=2求sin(x)cos(x)=? tan x=2,求 sin x * cos x=? 已知:tan x=2 求:sin x+cos x/sin x-cos x=? 若tanx=2,求:sin(tt/2+x)*cos(tt/2-x)*tan(-x+3tt)/sin(7tt-x)*tan(6tt-x)tt 表示 “派” 已知sin(π+x)=-1/2,求cos(2π-x),tan(x-7π) 已知sin x=15/17,x属于π/2,π,求tan(π/4+x) 证明:(tan^2)x-(sin^2)x=(tan^2)x(sin^2)x 求证tan^x-sin^2x=tan^2x*sin^2x 求证:(sin 2x /(1-cos 2x) )·(sin x /(1+sin x))=tan (π/4-x/2). sin(3π/2-2x)=3/5 求tan²x 证明tan^2x-sin^2x=tan^2 sin^2x sin(180+x)+cos(-x)/sin(180-x)+cos(360+x)=2 求tan(x+45） 求∫1／tan^2x+sin^2x dx 若x属于(0,π/2),且6 cosx=5 tan X,则sin X 求证：tan x/2=sin x/(1+cos x)