Lời giải của GV Vungoi.vn
Ta có:
$\left( {1 + \tan a} \right)\left( {1 + \tan b} \right)$\( = 1 + \tan a + \tan b + \tan a\tan b\) \( = 1 + \tan \left( {a + b} \right)\left( {1 - \tan a\tan b} \right) + \tan a\tan b\)
\( = 1 + \tan \left( {{{20}^0} + {{25}^0}} \right)\left( {1 - \tan {{20}^0}.\tan {{25}^0}} \right) + \tan {20^0}.\tan {25^0}\) \( = 1 + \tan {45^0}\left( {1 - \tan {{20}^0}\tan {{45}^0}} \right) + \tan {20^0}\tan {25^0}\) \( = 1 + 1 - \tan {20^0}\tan {25^0} + \tan {20^0}\tan {25^0} = 2\)
Đáp án cần chọn là: b
Cách khác:
$C = \left( {1 + \tan a} \right)\left( {1 + \tan b} \right) = \dfrac{{\left( {\cos a + \sin a} \right)}}{{\cos a}}.\dfrac{{\left( {\cos b + \sin b} \right)}}{{\cos b}} = \dfrac{{2\sin \left( {a + 45^\circ } \right)\sin \left( {b + 45^\circ } \right)}}{{\cos a\cos b}}$
\(C = \dfrac{{\cos \left( {a - b} \right) - \cos \left( {a + b + 90^\circ } \right)}}{{\dfrac{1}{2}\left[ {\cos \left( {a + b} \right) + \cos \left( {a - b} \right)} \right]}} = \dfrac{{\cos 5^\circ - \cos \left( {45^\circ + 90^\circ } \right)}}{{\dfrac{1}{2}\left( {\cos 45^\circ + \cos 5^\circ } \right)}} = \dfrac{{\cos 5^\circ + \sin 45^\circ }}{{\dfrac{1}{2}\left( {\cos 45^\circ + \cos 5^\circ } \right)}} = 2\).
