Cho α thoả mãn tanα = 3 và 0 < α < pi/2.Khi đó:

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

a) Đúng. \(\cot \alpha = \frac{1}{{\tan \alpha }} = \frac{1}{3}\).

b) Sai. Ta có \(A = \frac{1}{{{{\cos }^2}\alpha }} = 1 + {\tan ^2}\alpha = 10\).

c) Đúng. Do tanα = 3 nên cosα ≠ 0.

\(B = \frac{{\sin \alpha + 5\cos \alpha }}{{3\sin \alpha + 6\cos \alpha }} = \frac{{\frac{{\sin \alpha }}{{\cos \alpha }} + \frac{{5\cos \alpha }}{{\cos \alpha }}}}{{\frac{{3\sin \alpha }}{{\cos \alpha }} + \frac{{6\cos \alpha }}{{\cos \alpha }}}} = \frac{{\tan \alpha + 5}}{{3\tan \alpha + 6}} = \frac{{3 + 5}}{{3 \cdot 3 + 6}} = \frac{8}{{15}}\).

d) Đúng. C = \(\frac{{\sin 3\alpha \cdot \cos 2\alpha + \sin \alpha \cdot \cos 6\alpha }}{{\sin 4\alpha }}\)

= \(\frac{{\frac{1}{2}\left[ {\sin \left( {3\alpha + 2\alpha } \right) + \sin \left( {3\alpha - 2\alpha } \right)} \right] + \frac{1}{2}\left[ {\sin \left( {\alpha + 6\alpha } \right) - \sin \left( {\alpha - 6\alpha } \right)} \right]}}{{\sin 4\alpha }}\)

\( = \frac{{\frac{1}{2}\left( {\sin 5\alpha + {\mathop{\rm s}\nolimits} {\rm{in}}\alpha } \right) + \frac{1}{2}\left( {\sin 7\alpha - \sin 5\alpha } \right)}}{{\sin 4\alpha }} = \frac{{\frac{1}{2}\left( {\sin \alpha + \sin 7\alpha } \right)}}{{\sin 4\alpha }} = \frac{{{\mathop{\rm s}\nolimits} {\rm{in}}\alpha + \sin 7\alpha }}{{2\sin 4\alpha }}\)

= \(\frac{{\left( {\sin 7\alpha + {\mathop{\rm s}\nolimits} {\rm{in}}\alpha } \right) \cdot \cos 3\alpha }}{{2\sin 4\alpha \cdot \cos 3\alpha }} = \frac{{\left( {\sin 7\alpha + {\mathop{\rm s}\nolimits} {\rm{in}}\alpha } \right) \cdot \cos 3\alpha }}{{\sin 7\alpha + {\mathop{\rm s}\nolimits} {\rm{in}}\alpha }} = \cos 3\alpha \).