Cho \(\tan \alpha = 1\). Tính \(B = \frac{{{{\sin }^2}\alpha + 1}}{{2{{\cos }^2}\alpha - {{\sin }^2}\alpha }}\).

\(F = 1 - 2{\sin ^2}55^\circ  + 4{\cos ^2}60^\circ  - 2{\sin ^2}35^\circ  + \tan 55^\circ \tan 35^\circ \)\[ = 1 - 2{\sin ^2}\left( {90^\circ  - 35^\circ } \right) + 4{\cos ^2}60^\circ  - 2{\sin ^2}35^\circ  + \tan \left( {90^\circ  - 35^\circ } \right)\tan 35^\circ \]\( = 1 - 2{\cos ^2}35^\circ  + 4{\cos ^2}60^\circ  - 2{\sin ^2}35^\circ  + \cot 35^\circ  \cdot \tan 55^\circ \)\( = 1 - 2\left( {{{\sin }^2}35^\circ  + {{\cos }^2}35^\circ } \right) + 4{\cos ^2}60^\circ  + \tan 35^\circ \cot 35^\circ  = 1 - 2 \cdot 1 + 4 \cdot {\left( {\frac{1}{2}} \right)^2} + 1 = 1\).

Đáp án: 1.

Đáp án đúng là: B

Ta có \(\cos 45^\circ  + \sin 45^\circ \,{\rm{ = }}\frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} = \sqrt 2 \).