`~rai~`
\(\dfrac{1}{\sin^2x}+\dfrac{1}{\cos^2x}=\tan^2x+\cot^2x+2.\\\text{Xét VT=}\dfrac{1}{\sin^2x}+\dfrac{1}{\cos^2x}\\\quad\quad\quad=\dfrac{\sin^2x+\cos^2x}{\sin^2x}+\dfrac{\sin^2x+\cos^2x}{\cos^2x}\\\quad\quad\quad=\dfrac{\sin^2x}{\sin^2x}+\dfrac{\cos^2x}{\sin^2x}+\dfrac{\sin^2x}{\cos^2x}+\dfrac{\cos^2x}{\cos^2x}\\\quad\quad\quad=1+\cot^2x+\tan^2x+1\\\quad\quad\quad=\tan^2x+\cot^2x+2\\\quad\quad\quad=VP.\\\Rightarrow \dfrac{1}{\sin^2x}+\dfrac{1}{\cos^2x}=\tan^2x+\cot^2x+2.(đpcm)\\\text{Giải thích:Các công thức đã dùng để chứng minh:}\\+)\sin^2\alpha+\cos^2\alpha=1.\\+)\dfrac{\sin\alpha}{\cos\alpha}=\tan\alpha.\\+)\dfrac{\cos\alpha}{\sin\alpha}=\cot\alpha.\)
