Phần II. Trắc nghiệm đúng, sai

Cho \(\sin \alpha = \frac{3}{5},\left( {90^\circ < \alpha < 180^\circ } \right)\).

a) \(\cos \alpha  > 0\).

b) \({\cos ^2}\alpha  = \frac{{16}}{{25}}\).

c) \[\tan \left( {180^\circ  - \alpha } \right) =  - \frac{3}{4}\].

d) \[A = \frac{{\tan \alpha  - \cot \left( {180^\circ  - \alpha } \right)}}{{\sin \left( {90^\circ  - \alpha } \right)}} = \frac{{125}}{{48}}\].

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

Đáp án: 1.

Đáp án đúng là: D

\(A = \left( {\tan 1^\circ  \cdot \tan 89^\circ } \right) \cdot \left( {\tan 2^\circ  \cdot \tan 88^\circ } \right) \cdot ... \cdot \left( {\tan 44^\circ  \cdot \tan 46^\circ } \right) \cdot \tan 45^\circ \)

\[ = \left( {\tan 1^\circ  \cdot \cot 1^\circ } \right) \cdot \left( {\tan 2^\circ  \cdot \cot 2^\circ } \right) \cdot ... \cdot \left( {\tan 44^\circ  \cdot \cot 44^\circ } \right) \cdot \tan 45^\circ \]

\( = 1\).