sec+2与tan+2的关系

骆欣畏2409求∫1/(2+sinx)dx的不定积分 -
封沫贸18575176272 ______ ∫1/(2+sinx)dx=2√3/3*arctan{[2√3tan(x/2)+√3]/3}+C.C为常数. 2+sinx=2sin(x/2)^2+2cos(x/2)^2+2sin(x/2)cos(x/2) dx/(2+sinx)=sec(x/2)^2dx/[2+2tan(x/2)^2+2tan(x/2)] =d(tan(x/2))/[1+tan(x/2)+tan(x/2)^2] 令u=tan(x/2) 原积分=∫du/(1+u+u^2) =∫d(u+...

骆欣畏2409如果简化: (tan^2 x / sec x + 1) +1 -
封沫贸18575176272 ______ 1+tan^2=secx^2得tan^2=secx^2-1 [tan^2 x / (sec x + 1) ]+1=[(secx^2-1)/(secx+1)]+1=[(secx-1)(secx+1)/(secx+1)]+1=secx-1+1=secx

骆欣畏2409请帮忙求证:(1+sec2θ)/tan2θ=cotθ -
封沫贸18575176272 ______ 左边=[1+(1/cos2θ)]/[sinθ/cos2θ]=[(1+cos2θ)/(cos2θ)]*[cos2θ/sin2θ]=cos2θ/sin2θ=cot2θ=右边 则:(1+sec2θ)/tan2θ=cot2θ

骆欣畏2409...(1 - tanα)=2001 求证:sec2α+tan2α=2001如果您没有好思路,我这里有参考答案,但其中有一步我不太懂:【证明】:sec2α+tan2α=(1+sin2α)/cos2α=[1 - cos(π... -
封沫贸18575176272 ______[答案] 半角公式:tan(α/2)=sinα/(1+cosα)=(1-cosα)/sinα

骆欣畏2409求证(1 - tan^a)^=(sec^a - 2tana)(sec^a+2tana).要过程谢谢. -
封沫贸18575176272 ______ sec^a=1+tan^a sec^a-2tana=tan^a-2tana+1=(1-tana)^ sec^a+2tana=tan^a+2tana+1=(1+tana)^ ∴(sec^a-2tana)(sec^a+2tana)=(1-tana)^(1+tana)^=(1-tan^a)^

骆欣畏2409sec^2(x/2)/2tan(x/2)等于1/sinx,麻烦高手帮我把详细步骤写出来!! -
封沫贸18575176272 ______ sec^2(x/2)/[2tan(x/2)] =[1/cos²(x/2)]÷[2sin(x/2)/cos(x/2)] =[1/cos²(x/2)]*cos(x/2)/[2sin(x/2)] =1/[2sin(x/2)*cos(x/2)] =1/sinx

骆欣畏2409已知x属于【 - 派/6,派/4】,求函数y=sec^2x+tanx+2的最值 -
封沫贸18575176272 ______ 令t=tanx, 则t的取值范围是[-√3/3, 1] 则y=t²+1+t+2=(t+1/2)²+11/4 当t=-1/2时,y取最小值11/4 当t=1时,y取最大值5

骆欣畏2409解方程:(sec^2)x=1+tanx -
封沫贸18575176272 ______ sec^2 x=1+tanx=1+tan^2 xtanx=tan^2 xtanx=1 or 0x=π/4+kπ or kπ , k为任意整数

骆欣畏2409y=sec^2θ+2tanθ+1 最大值 -
封沫贸18575176272 ______ 解: sec^2θ=1/cos^2θ,代入化简有y=(1+2sinθcosθ)/cos^2θ+1,将sin^2θ+cos^2θ=1代入有Y=1+(1+tanθ)^2(好像条件不足啊)

骆欣畏2409求该函数的导数 tan(x^2+1) sec2x -
封沫贸18575176272 ______ (2*(1+tan(x^2+1)^2))*x*sec(2*x)+2*tan(x^2+1)*sec(2*x)*tan(2*x)