求证[tanθ·(1-sinθ)]/1+cosθ=[cotθ·(1-cosθ)]/1+sinθ

求证 tanθ(1+sinθ )+sinθ /tanθ (1+sinθ )-sinθ =tanθ+sinθ/tanθsinθ 求证sinθ（1+tanθ)+cosθ(1+1/tanθ)=1/sinθ+1/cosθ 求证[tanθ·(1-sinθ)]/1+cosθ=[cotθ·(1-cosθ)]/1+sinθ 求证2(cosθ -sinθ )/1+sinθ +cosθ =tan(π/4-θ /2)-tanθ /2 求证：2（cos θ -sin θ )/(1+sinθ +cosθ)=tan(∏/4- θ /2)-tan(θ /2) 求证：tan²θ-sin²θ=tan²θ·sin²θ 已知(sin^2α/sin^2β)+cos^2αcos^2θ=1,求证tan^2α=sin^2θtan^2β 已知sin²α/sin²β+cos²αcosθ=1,求证tan²α=sin²θtan²β 已知tan^2θ=2tan^2a+1,求证:cos2θ+sin^2a=0 tan²α=2tan²θ+1求证cos2α+sin²θ=0 已知tan^2θ=2tan^2a+1,求证:cos2θ+sin^2a=0 求证1/sin2θ + 1/tan2θ +1/sinθ =1/tanθ/2 求证(sinθ+cosθ-1)(sinθ-cosθ+1)/sin2θ=tanθ/2 求证 (sinθ+cosθ-1)(sinθ-cosθ+1)) /sin2θ=tanθ/2 求证：sin³θ（1+cotθ）+cos³θ（1+tanθ）=sinθ+cosθ 求证：[2sin(θ-3π/2)cos(θ+π/2)-1]/1-2sin^2 θ=[tan(9π+θ)+1]/tanθ-1 tan^2θ =2tan^2Ф+1 求证 cos2θ +sin^2Ф=0如题.sin^2a/(1-sin^2a)=2sin^2⊙/(1-sin^2⊙)+1sin^2a/(1-sin^2a+sin^2a)=(sin^2⊙+1)/(1-sin^2⊙+sin^2⊙+1)这是怎么来的。 求证tan^2θ-sin^2θ=tan^2θsin^2θ