求证 (1/sin^2 x) + (1/cos^2 x) - （1/tan^2 x）= 2 + (tan^2 x)怎么证明啊……?

求证：tan x/2=sin x/(1+cos x) 求证：1-sin(x)=cos(x)*tan(x/2) 求证：(sin x+cos x)(1-tan x)=(2sin x)/(tan2x) 求证:(sin 2x /(1-cos 2x) )·(sin x /(1+sin x))=tan (π/4-x/2). 求证 cos^8(x)-sin^8(x)=cos2x【1-1/2sin^2（2x)】 求证sin^4x+cos^4x=1-2sin^2xcos^2x 三角等式求证：cos^6x+sin^6x=1-3sin^2x+3sin^4x 求证（cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2x 1-cosx=2sin²(x/2)求证 求证:Sin^2 x / (sinx-cosx) - (sin x + cos x)/(ta求证:Sin^2 x / (sinx-cosx) - (sin x + cos x)/(tan^2-1) = sin x + cos x 求证(3-sin^4 x-cos^4 x)/2cos^2 x=1+tan^2 x+sin^2 x 求证 sin的4次方x减去cos4次方a等于2sin平方a-1 求证 1+2sinxcosx/sin^2x-cos^2x=sin^2x-cos^2x/1-2sinxcosx 求证：1-cos^4x-sin^4x/1-cos^6x-sin^6x=2/3 求证（cos^2x-sin^2x)(cos^4x+sin^4x）+1/4sin2xsin4x=cos2x 求证sin^2x+sin^2y-sin^2x*sin^2y+cos^2x*cos^2y=1 求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)] 求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)]