作业帮求证：tan(x/2+派/4)+tan(x/2-派/4)=2tanx{[tan(x/2)+1]^2-[1-tan(x/2)]^2}=4tan(x/2)具体怎么得到
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![求证：tan(x/2+派/4)+tan(x/2-派/4)=2tanx{[tan(x/2)+1]^2-[1-tan(x/2)]^2}=4tan(x/2)具体怎么得到](/uploads/image/z/8573740-52-0.jpg?t=%E6%B1%82%E8%AF%81%EF%BC%9Atan%28x%2F2%2B%E6%B4%BE%2F4%29%2Btan%28x%2F2-%E6%B4%BE%2F4%29%3D2tanx%7B%5Btan%28x%2F2%29%2B1%5D%5E2-%5B1-tan%28x%2F2%29%5D%5E2%7D%3D4tan%28x%2F2%29%E5%85%B7%E4%BD%93%E6%80%8E%E4%B9%88%E5%BE%97%E5%88%B0)
求证：tan(x/2+派/4)+tan(x/2-派/4)=2tanx{[tan(x/2)+1]^2-[1-tan(x/2)]^2}=4tan(x/2)具体怎么得到
求证：tan(x/2+派/4)+tan(x/2-派/4)=2tanx
{[tan(x/2)+1]^2-[1-tan(x/2)]^2}=4tan(x/2)具体怎么得到
求证：tan(x/2+派/4)+tan(x/2-派/4)=2tanx{[tan(x/2)+1]^2-[1-tan(x/2)]^2}=4tan(x/2)具体怎么得到
设,tan(x/2)=a
（a+1)^2-(1-a)^2=a^2+2a+1-（a^2-2a+1）=4a=4tan(x/2)
直接按完全平方公式展开就得到了
tan(x/2+派/4)+tan(x/2-派/4)=[sin(x/2+派/4)*cos(x/2-派/4)+cos(x/2+派/4)*sin(x/2-派/4)]/[cos(x/2-派/4)cos(x/2+派/4)]=sinx / [1/2*（cosx+cos派/2)]=sinx/[1/2*cosx]=2tanx
求证：tan(x/2+派/4)+tan(x/2-派/4)=2tanx
求证：tan(x/2+派/4)+tan(x/2-派/4)=2tanx{[tan(x/2)+1]^2-[1-tan(x/2)]^2}=4tan(x/2)具体怎么得到
y=tan(2x-派/4)
求证恒等式：(1+sinx)/cosx=tan(派/4+x/2);
化简:cos(x-派)tan(x-2派)tan(2派-x)/sin(派+X)
tan(-5派/4)=?tan(-2派/3)=?
tan x + 派=2 x=?
已知sin(2x+y)=5siny,(y不等于K派),求证;3tan=2tan(x+y)
化简:sin(2派-x)tan(派+x)cot(-x-派)/tan(3派-x)cos(派-x)
函数 tan（2x-4分之派）的定义域
y=tan（2x+派/4）的定义域
lim x趋近于派/2 tan-5/sec+4
tanx=2,则tan(派/4+x)=?
tanx=2,则tan(派/4+x)=?
已知tan（x＋5派／4 ）＝2 求tan xr t
tan(2分之x+4分之派)乘tan(2分之x-4分 之派)=?
tan(派/2+a)
求证tan(x+y)*tan(x-y)=tan^2x-tan^2y/1-tan^2xtan^2y