Giải bài 9 trang 161 – SGK môn Đại số lớp 10

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$\begin{align} a)\,&4\left( \cos {{24}^{o}}+\cos {{48}^{o}}-\cos {{84}^{o}}-\cos {{12}^{o}} \right) \\ & =4\left[ \left( \cos {{24}^{o}}-\cos {{84}^{o}} \right)+\left( \cos {{48}^{o}}-\cos {{12}^{o}} \right) \right] \\ & =8\sin {{54}^{o}}\sin {{30}^{o}}+8\sin {{30}^{o}}\sin {{18}^{o}} \\ & =4\left( \sin {{54}^{o}}+\sin {{18}^{o}} \right) \\ & =8\cos {{36}^{o}}\sin {{18}^{o}} \\ & =\dfrac{8\cos {{36}^{o}}\sin {{18}^{o}}\cos {{18}^{o}}}{\cos {{18}^{o}}} \\ & =\dfrac{4\cos {{36}^{o}}\sin {{36}^{o}}}{\sin {{72}^{o}}}=2 \\ \end{align}$
$\begin{align} b)& \,96\sqrt{3}\sin \dfrac{\pi }{48}\cos \dfrac{\pi }{48}\cos \dfrac{\pi }{24}\cos \dfrac{\pi }{12}\cos \dfrac{\pi }{6} \\ & =48\sqrt{3}\sin \dfrac{\pi }{24}\cos \dfrac{\pi }{24}\cos \dfrac{\pi }{12}\cos \dfrac{\pi }{6} \\ & =24\sqrt{3}\sin \dfrac{\pi }{12}\cos \dfrac{\pi }{12}\cos \dfrac{\pi }{6} \\ & =12\sqrt{3}\sin \dfrac{\pi }{6}\cos \dfrac{\pi }{6} \\ & =6\sqrt{3}\sin \dfrac{\pi }{3}=9 \\ \end{align}$ 
$\begin{align} c)& \,\tan {{9}^{o}}-\tan {{63}^{o}}+\tan {{81}^{o}}-\tan {{27}^{o}} \\ & =\tan {{9}^{o}}+\cot {{9}^{o}}-\left( \tan {{27}^{o}}+\cot {{27}^{o}} \right) \\ & =\dfrac{\sin {{9}^{o}}}{\cos {{9}^{o}}}+\dfrac{\cos {{9}^{o}}}{\sin {{9}^{o}}}-\left( \dfrac{\sin {{27}^{o}}}{\cos {{27}^{o}}}+\dfrac{\cos {{27}^{o}}}{\sin {{27}^{o}}} \right) \\ & =\dfrac{1}{\sin {{9}^{o}}\cos {{9}^{o}}}-\dfrac{1}{\sin {{27}^{o}}\cos {{27}^{o}}} \\ & =\dfrac{2}{\sin {{18}^{o}}}-\dfrac{2}{\sin {{54}^{o}}} \\ & =\dfrac{4\cos {{36}^{o}}\sin {{18}^{o}}}{\sin {{18}^{o}}\sin {{54}^{o}}} \\ & =4 \\ \end{align}$