PHẦN II. TRẮC NGHIỆM ĐÚNG – SAI

 Cho \(0 < a < \frac{\pi }{2};\frac{\pi }{2} < b < \pi \) và tana = 3; tanb = −2.

a) tan(a + π) = −3.

b) tan(a + b) = −1.

c) cot(a − b) = 1.

d) \(\sin \left( {a - b} \right) = - \frac{{\sqrt 2 }}{2}\).

a) tan(a + π) = tana = 3.

b) \(\tan \left( {a + b} \right) = \frac{{\tan a + \tan b}}{{1 - \tan a\tan b}} = \frac{{3 + \left( { - 2} \right)}}{{1 - 3.\left( { - 2} \right)}} = \frac{1}{7}\).

c) \(\cot \left( {a - b} \right) = \frac{1}{{\tan \left( {a - b} \right)}} = \frac{1}{{\frac{{\tan a - \tan b}}{{1 + \tan a.\tan b}}}} =  - 1\).

d) Ta có \(0 < a < \frac{\pi }{2}\); tana = 3 Þ cosa > 0.

Có \(1 + {\tan ^2}a = \frac{1}{{{{\cos }^2}a}} \Rightarrow \cos a = \frac{{\sqrt {10} }}{{10}}\); \(\sin a = \tan a.\cos a = \frac{{3\sqrt {10} }}{{10}}\).

Ta có \(\frac{\pi }{2} < b < \pi ;\tan b =  - 2 \Rightarrow \cos b < 0\).

\(1 + {\tan ^2}b = \frac{1}{{{{\cos }^2}b}} \Rightarrow \cos b =  - \frac{{\sqrt 5 }}{5}\); \(\sin b = \tan b.\cos b = \frac{{2\sqrt 5 }}{5}\).

Do đó \(\sin \left( {a - b} \right) = \sin a.\cos b - \cos a.\sin b = \frac{{3\sqrt {10} }}{{10}}.\left( {\frac{{ - \sqrt 5 }}{5}} \right) - \frac{{\sqrt {10} }}{{10}}.\frac{{2\sqrt 5 }}{5} =  - \frac{{\sqrt 2 }}{2}\).

Đáp án: a) Sai;   b) Sai;   c) Sai;   d) Đúng.