Chọn B.

Ta có: A = (tanx + cotx)² - ( tanx - cotx)²

= tan²x +  2tanx.cot x + cot²x - ( tan²x - 2tanx.cotx + cot²x)

= 4tanx.cotx = 4.

\(A=sin^3x\cdot\left(1+\dfrac{cosx}{sinx}\right)+cos^3x\left(1+\dfrac{sinx}{cosx}\right)\)

\(=sin^2x\left(sinx+cosx\right)+cos^2x\left(cosx+sinx\right)\)
=cosx+sinx

\(\cot x-\tan x\)

\(=\frac{cosx}{\sin x}-\frac{\sin x}{cosx}=\frac{cos^2x-\sin^2x}{\sin x\cdot cosx}=\frac{2\cdot cos2x}{\sin2x}=2\cdot\cot2x\)

=>ĐPCM

cot a-tan a=2*cot2a

=>tan a=cota-2*cot2a

\(2\cdot\tan2a=2\left(\cot2a-2\cot4a\right)=2\cdot\cot2a-4\cdot\cot4a\)

...

\(2^{n}\cdot\left(\tan2^{n}\cdot a\right)=2^{n}\cdot\cot\left(2^{n}\cdot a\right)-2^{n+1}\cdot\cot\left(2^{n+1}\cdot a\right)\)

Do đó: S=\(\left(\cot a-2\cdot\cot2a\right)+\left(2\cdot\cot2a-4\cdot\cot4a\right)+\cdots+\left\lbrack2_{}^{n}\cdot\cot\left(2^{n}\cdot a\right)-2^{n+1}\cdot\cot\left(2^{n+1}\cdot a\right)\right\rbrack\)

\(=\cot a-2^{n+1}\cdot\cot\left(2^{n+1}\cdot a\right)\)

tích mình với

ai tích mình

mình tích lại

thanks

Sin 1 slot xíu nữa làm

\(\left(tanx-cotx\right)^2=9\Rightarrow tan^2x-2.tanx.cotx+cot^2x=9\)

\(\Rightarrow tan^2x+cot^2x=11\)

\(\left(tanx+cotx\right)^2=tan^2x+cot^2x+2.tanx.cotx=11+2=13\)

\(\Rightarrow tanx+cotx=\pm\sqrt{13}\)

\(tan^4x-cot^4x=\left(tan^2x+cot^2x\right)\left(tan^2x-cot^2x\right)\)

\(=11\left(tanx+cotx\right)\left(tanx-cotx\right)=\pm33\sqrt{13}\)

Chắc bạn ghi sai đề, là \(tanx+cotx=m\) mới đúng (vì \(tanx.cotx=1\))

\(\Rightarrow\left(tanx+cotx\right)^2=m^2\)

\(\Leftrightarrow\left(tanx-cotx\right)^2+4tanx.cotx=m^2\)

\(\Leftrightarrow\left(tanx-cotx\right)^2=m^2-4\)

\(\Rightarrow\left|tanx-cotx\right|=\sqrt{m^2-4}\)