Đáp án:
$\begin{array}{l}
a)B = 0\\
b)D = \sin 2x\\
c)F = \cos x\\
d)H = \tan a
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
a)B = \dfrac{{\sin \left( {a - b} \right)}}{{\cos a\cos b}} + \dfrac{{\sin \left( {b - c} \right)}}{{\cos b\cos c}} + \dfrac{{\sin \left( {c - a} \right)}}{{\cos c\cos a}}\\
= \dfrac{{\sin a\cos b - \sin b\cos a}}{{\cos a\cos b}} + \dfrac{{\sin b\cos c - \sin c\cos b}}{{\cos b\cos c}} + \dfrac{{\sin c\cos a - \sin a\cos c}}{{\cos c\cos a}}\\
= \dfrac{{\sin a}}{{\cos a}} - \dfrac{{\sin b}}{{\cos b}} + \dfrac{{\sin b}}{{\cos b}} - \dfrac{{\sin c}}{{\cos c}} + \dfrac{{\sin c}}{{\cos c}} - \dfrac{{\sin a}}{{\cos a}}\\
= 0\\
b)D = \sin \left( {x + y} \right)\cos \left( {x - y} \right) + \sin \left( {x - y} \right)\cos \left( {x + y} \right)\\
= \sin \left( {x + y + x - y} \right)\\
= \sin 2x\\
c)F = \sin \left( {x + {{10}^0}} \right)\cos \left( {2x - {{80}^0}} \right) + \sin \left( {x + {{100}^0}} \right)\cos \left( {2x + {{10}^0}} \right)\\
= \sin \left( {x + {{10}^0}} \right)\cos \left( {2x - {{80}^0}} \right) + \cos \left( { - x - {{10}^0}} \right)\sin \left( { - 2x + {{80}^0}} \right)\\
= \sin \left( {x + {{10}^0}} \right)\cos \left( {2x - {{80}^0}} \right) - \sin \left( {2x - {{80}^0}} \right)\cos \left( {x + {{10}^0}} \right)\\
= \sin \left( {x + {{10}^0} - \left( {2x - {{80}^0}} \right)} \right)\\
= \sin \left( {{{90}^0} - x} \right)\\
= \cos x\\
d)H = \dfrac{{2\sin \left( {a + b} \right)}}{{\cos \left( {a + b} \right) + \cos \left( {a - b} \right)}} - \tan b\\
= \dfrac{{2\left( {\sin a\cos b + \sin b\cos a} \right)}}{{2\cos a\cos b}} - \dfrac{{\sin b}}{{\cos b}}\\
= \dfrac{{\sin a}}{{\cos a}} + \dfrac{{\sin b}}{{\cos b}} - \dfrac{{\sin b}}{{\cos b}}\\
= \dfrac{{\sin a}}{{\cos a}}\\
= \tan a
\end{array}$
