Ta có các công thức cơ bản sau: \(cos\left(90^0+x\right)=-sinx;sin\left(90^0-x\right)=cosx\)

\(cot\left(90^0-x\right)=tanx;tan\left(90^0+x\right)=-cotx\)

Thay vào bài toán:

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

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

`(x/(x+1))^2+(x/(x-1))^2=90(x ne -1,1)`

`<=>x^2/(x+1)^2+x^2/(x-1)^2=90`

`<=>x^2(x-1)^2+x^2(x-1)^2=90(x^2-1)^2`

`<=>x^2(2x^2+2)=90(x^4-2x^2+1)`

`<=>2x^4+2x^2=90x^4-180x^2+90`

`<=>88x^4-182x^2+90=0`

`<=>88x^4-110x^2-72x^2+90=0`

`<=>22x^2(4x^2-5)-18(4x^2-5)=0`

`<=>(4x^2-5)(22x^2-18)=0`

`<=>(4x^2-5)(11x^2-9)=0`

`<=>` $\left[ \begin{array}{l}4x^2=5\\11x^2=9\end{array} \right.$

`<=>` $\left[ \begin{array}{l}x=\sqrt{\dfrac{5}{4}}\\x=-\sqrt{\dfrac{5}{4}}\\x=\sqrt{\dfrac{9}{11}}\\x=-\sqrt{\dfrac{9}{11}}\end{array} \right.$

Vậy `S={\sqrt{9/11},-\sqrt{9/11},\sqrt{5/4},-\sqrt{5/4}}`

\(\left(\dfrac{x}{x+1}\right)^2+\left(\dfrac{x}{x-1}\right)^2=90\)

\(\Leftrightarrow\dfrac{x^2}{\left(x+1\right)^2}+\dfrac{x^2}{\left(x-1\right)^2}=90\)

\(\Leftrightarrow\dfrac{x^2\left(x-1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}+\dfrac{x^2\left(x+1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}=90\)

\(\Leftrightarrow\dfrac{x^2\left(x-1\right)^2+x^2\left(x+1\right)^2-90\left(x-1\right)^2\left(x+1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}=0\)

\(\Rightarrow x^2\left(x^2-2x+1\right)+x^2\left(x^2+2x+1\right)-90\left(x^2-1\right)^2=0\)

\(\Leftrightarrow x^4-2x^3+x^2+x^4+2x^3+x^2-90x^4+90x^2-90=0\)

\(\Leftrightarrow-88x^4+92x^2-90=0\)

\(\left(\dfrac{x}{x+1}\right)^2+\left(\dfrac{x}{x-1}\right)^2+\dfrac{2x^2}{x^2-1}-\dfrac{2x^2}{x^2-1}=90\)

\(\Leftrightarrow\left(\dfrac{x}{x+1}+\dfrac{x}{x-1}\right)^2-\dfrac{2x^2}{x^2-1}=90\)

\(\Leftrightarrow\left(\dfrac{2x^2}{x^2-1}\right)^2-\dfrac{2x^2}{x^2-1}-90=0\)

Đặt \(\dfrac{2x^2}{x^2-1}=t\Rightarrow t^2-t-90=0\Rightarrow\left[{}\begin{matrix}t=10\\t=-9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2x^2}{x^2-1}=10\\\dfrac{2x^2}{x^2-1}=-9\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}4x^2=5\\11x^2=9\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(\dfrac{5\left|x+1\right|}{2}=\dfrac{90}{\left|x+1\right|}\left(x\ne-1\right)\\ \Rightarrow5\left|x+1\right|^2=90\cdot2=180\\ \Rightarrow\left|x+1\right|^2=36\\ \Rightarrow\left|x+1\right|=6\left(\left|x+1\right|>0\right)\\ \Rightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)

Ta có \(\left(x-\frac{1}{2}\right)+\left(x-\frac{1}{6}\right)+\left(x-\frac{1}{12}\right)+...+\left(x-\frac{1}{90}\right)=1\)

\(\Rightarrow\left(x+x+x+...+x\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)=1\)

\(\Rightarrow9x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\right)=1\)

\(\Rightarrow9x-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)=1\)

\(\Rightarrow9x-\left(1-\frac{1}{10}\right)=1\)

\(\Rightarrow9x-\frac{9}{10}=1\)

\(\Rightarrow9x=\frac{19}{10}\)

\(\Rightarrow x=\frac{19}{10}\)