Đáp án:
-11
Giải thích các bước giải:
Từ giả thiết ta có
$2\left( {\cos \alpha \cos \beta - \sin \alpha \sin \beta } \right) = - \cos \alpha \cos \beta \Rightarrow \tan \beta = \dfrac{3}{{2\tan \alpha }}$ .
Khi đó ta có:
$A = \dfrac{{1 + {{\tan }^2}\alpha }}{{2{{\tan }^2}\alpha + 3}} + \dfrac{{1 + {{\tan }^2}\beta }}{{2{{\tan }^2}\beta + 3}} = \dfrac{{1 + {{\tan }^2}\alpha }}{{2{{\tan }^2}\alpha + 3}} + \dfrac{{1 + \dfrac{9}{{4{{\tan }^2}\alpha }}}}{{2\dfrac{9}{{4{{\tan }^2}\alpha }} + 3}}$
$A = \dfrac{{1 + {{\tan }^2}\alpha }}{{2{{\tan }^2}\alpha + 3}} + \dfrac{{4{{\tan }^2}\alpha + 9}}{{6\left( {2{{\tan }^2}\alpha + 3} \right)}} = \dfrac{{10{{\tan }^2}\alpha + 15}}{{6\left( {2{{\tan }^2}\alpha + 3} \right)}} = \dfrac{5}{6}$
Vậy ${{a}^{2}}-{{b}^{2}}={{5}^{2}}-{{6}^{2}}=-11$