mn giúp e với ạ em cảm ơn ạ

Đáp án:

$A=1$

Giải thích các bước giải:

$\tan \alpha=2\\ \tan^2\alpha+1=\dfrac{\sin^2\alpha}{\cos^2\alpha}+1=\dfrac{\sin^2\alpha+\cos^2\alpha}{\cos^2\alpha}=\dfrac{1}{\cos^2\alpha}\\ \Rightarrow \cos^2\alpha=\dfrac{1}{\tan^2\alpha+1}\\ \sin^2\alpha=1- \cos^2\alpha=1-\dfrac{1}{\tan^2\alpha+1}=\dfrac{\tan^2\alpha}{\tan^2\alpha+1}\\ \sin^2\alpha\cos^2\alpha=\dfrac{\tan^2\alpha}{(\tan^2\alpha+1)^2}\\ \tan \alpha=2>0 \Rightarrow \sin\alpha\cos\alpha>0 \Rightarrow \sin\alpha\cos\alpha=\sqrt{\dfrac{\tan^2\alpha}{(\tan^2\alpha+1)^2}}\\ A=\dfrac{\tan^2\alpha}{\tan^2\alpha+1}+2\sqrt{\dfrac{\tan^2\alpha}{(\tan^2\alpha+1)^2}}-\dfrac{3}{\tan^2\alpha+1}\\ =1$