\(E=\frac{\frac{1}{sin^2x}}{1-\frac{cosx}{sinx}+\frac{cos^2x}{sin^2x}}=\frac{1+cot^2x}{1-cotx+cot^2x}=\frac{1+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{4}}=...\)

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

\(B=\left(tanx+cotx\right)^3-3tanx.cotx\left(tanx+cotx\right)=2^3-3.1.2=2\)

a1.

$\cot (2x+\frac{\pi}{3})=-\sqrt{3}=\cot \frac{-\pi}{6}$

$\Rightarrow 2x+\frac{\pi}{3}=\frac{-\pi}{6}+k\pi$ với $k$ nguyên

$\Leftrightarrow x=\frac{-\pi}{4}+\frac{k}{2}\pi$ với $k$ nguyên

a2. ĐKXĐ:...............

$\cot (3x-10^0)=\frac{1}{\cot 2x}=\tan 2x$

$\Leftrightarrow \cot (3x-\frac{\pi}{18})=\cot (\frac{\pi}{2}-2x)$

$\Rightarrow 3x-\frac{\pi}{18}=\frac{\pi}{2}-2x+k\pi$ với $k$ nguyên

$\Leftrightarrow x=\frac{\pi}{9}+\frac{k}{5}\pi$ với $k$ nguyên.

a3. ĐKXĐ:........

$\cot (\frac{\pi}{4}-2x)-\tan x=0$

$\Leftrightarrow \cot (\frac{\pi}{4}-2x)=\tan x=\cot (\frac{\pi}{2}-x)$

$\Rightarrow \frac{\pi}{4}-2x=\frac{\pi}{2}-x+k\pi$ với $k$ nguyên

$\Leftrightarrow x=-\frac{\pi}{4}+k\pi$ với $k$ nguyên.

a4. ĐKXĐ:.....

$\cot (\frac{\pi}{6}+3x)+\tan (x-\frac{\pi}{18})=0$

$\Leftrightarrow \cot (\frac{\pi}{6}+3x)=-\tan (x-\frac{\pi}{18})=\tan (\frac{\pi}{18}-x)$

$=\cot (x+\frac{4\pi}{9})$

$\Rightarrow \frac{\pi}{6}+3x=x+\frac{4\pi}{9}+k\pi$ với $k$ nguyên

$\Rightarrow x=\frac{5}{36}\pi + \frac{k}{2}\pi$ với $k$ nguyên. 

\(\frac{tan^3x}{sin^2x}-\frac{1}{sinx.cosx}+\frac{cot^3x}{cos^2x}=tan^3x\left(1+cot^2x\right)-\frac{1}{sinx.cosx}+cot^3x\left(1+tan^2x\right)\)

\(=tan^3x+tanx+cot^3x+cotx-\frac{1}{sinx.cosx}\)

\(=tan^3x+cot^3x+\frac{sinx}{cosx}+\frac{cosx}{sinx}-\frac{1}{sinx.cosx}\)

\(=tan^3x+cot^3x+\frac{sin^2x+cos^2x}{sinx.cosx}-\frac{1}{sinx.cosx}\)

\(=tan^3x+cot^3x\)

tích đúng mình làm cho

mình không hiểu 

Câu 1 đề sai, chắc chắn 1 trong 2 cái \(cot^2x\) phải có 1 cái là \(cos^2x\)

2.

\(\dfrac{1-sinx}{cosx}-\dfrac{cosx}{1+sinx}=\dfrac{\left(1-sinx\right)\left(1+sinx\right)-cos^2x}{cosx\left(1+sinx\right)}=\dfrac{1-sin^2x-cos^2x}{cosx\left(1+sinx\right)}\)

\(=\dfrac{1-\left(sin^2x+cos^2x\right)}{cosx\left(1+sinx\right)}=\dfrac{1-1}{cosx\left(1+sinx\right)}=0\)

3.

\(\dfrac{tanx}{sinx}-\dfrac{sinx}{cotx}=\dfrac{tanx.cotx-sin^2x}{sinx.cotx}=\dfrac{1-sin^2x}{sinx.\dfrac{cosx}{sinx}}=\dfrac{cos^2x}{cosx}=cosx\)

4.

\(\dfrac{tanx}{1-tan^2x}.\dfrac{cot^2x-1}{cotx}=\dfrac{tanx}{1-tan^2x}.\dfrac{\dfrac{1}{tan^2x}-1}{\dfrac{1}{tanx}}=\dfrac{tanx}{1-tan^2x}.\dfrac{1-tan^2x}{tanx}=1\)

5.

\(\dfrac{1+sin^2x}{1-sin^2x}=\dfrac{1+sin^2x}{cos^2x}=\dfrac{1}{cos^2x}+tan^2x=\dfrac{sin^2x+cos^2x}{cos^2x}+tan^2x\)

\(=tan^2x+1+tan^2x=1+2tan^2x\)

Ta có : \(tanx+cotx=m\)

\(\Rightarrow tan^2x+2tanx.cotx+cot^2x=m^2\)

\(\Rightarrow tan^2x+cot^2x=m^2-2tanx.cotx=m^2-2.1=m^2-2\)

Ta lại có : \(A=\left(tanx+cotx\right)\left(tan^2x-tanx.cotx+cot^2x\right)\)

\(=m\left(m^2-2-1\right)=m\left(m^2-3\right)=m^3-3m\)

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