\(\dfrac{1+\sin^2\alpha}{1-\sin^2\alpha}=1+2\tan^2\alpha\\\text{Xét VT=}\dfrac{1+\sin^2\alpha}{1-\sin^2\alpha}\\\quad\quad\quad=\dfrac{\sin^2\alpha+\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}\\\quad\quad\quad=\dfrac{\cos^2\alpha+2\sin^2\alpha}{\cos^2\alpha}\\\quad\quad\quad=\dfrac{\cos^2\alpha}{\cos^2\alpha}+\dfrac{2\sin^2\alpha}{\cos^2\alpha}\\\quad\quad\quad=1+2\tan^2\alpha\\\quad\quad\quad=VP.\\\Rightarrow \dfrac{1+\sin^2\alpha}{1-\sin^2\alpha}=1+2\tan^2\alpha.(đpcm)\\\text{Giải thích:Những công thức đã áp dụng để chứng minh:}\\+)\sin^2\alpha+\cos^2\alpha=1\\+)\dfrac{\sin\alpha}{\cos\alpha}=\tan\alpha.\)
