cho tam giác ABC vuông tại A. Biết tg=2. Tính $\frac{sin^2B+2cos^2B+1}{sin^2B-cos^2B+2}$

Đáp án: $\dfrac{11}{13}$

Giải thích các bước giải:

Ta có:

$A=\dfrac{\sin^2B+2\cos^2B+1}{\sin^2B-\cos^2B+2}$

$\to A=\dfrac{\sin^2B+2\cos^2B+(\sin^2B+\cos^2B)}{\sin^2B-\cos^2B+2(\sin^2B+\cos^2B)}$

$\to A=\dfrac{2\sin^2B+3\cos^2B}{3\sin^2B+\cos^2B}$

$\to A=\dfrac{2\cdot\dfrac{\sin^2B}{\cos^2B}+3}{3\cdot \dfrac{\sin^2B}{\cos^2B}+1}$

$\to A=\dfrac{2\cdot\tan^2B+3}{3\cdot \tan^2B+1}$

$\to A=\dfrac{2\cdot 2^2+3}{3\cdot 2^2+1}$

$\to A=\dfrac{11}{13}$