1)$\begin{array}{l} \cos {12^o} + \cos {18^o} - 4\cos {15^o}.\cos {21^o}.\cos {24^o}\\ = 2\cos {15^o}.\cos {3^o} - 4\cos {15^o}.\dfrac{1}{2}\left( {\cos \left( {{{21}^o} + {{24}^o}} \right) + \cos \left( {{{24}^o} - {{21}^o}} \right)} \right)\\ = 2\cos {15^o}.\cos {3^o} - 2\cos {15^o}\left( {\cos {{45}^o} + \cos {3^o}} \right)\\ = 2\cos {15^o}\left( {\cos {3^o} - \cos {{45}^o} - \cos {3^o}} \right)\\ = - 2\cos {15^o}.\cos {45^o} = - 2.\dfrac{1}{2}.\left( {\cos {{60}^o} + \cos {{30}^o}} \right) = \dfrac{{ - \left( {1 + \sqrt 3 } \right)}}{2} \end{array}$
$\begin{array}{l} \left\{ \begin{array}{l} {10^o} = A\\ {25^o} = B\\ {55^o} = C \end{array} \right. \to A + B + C = \dfrac{\pi }{2}\\ \to \tan \left( {A + B} \right) = \tan \left( {\dfrac{\pi }{2} - C} \right)\\ \to \tan \left( {A + B} \right) = \cot C = \dfrac{1}{{\tan C}}\\ \to \dfrac{{\tan A + \tan B}}{{1 - \tan A.\tan B}} = \dfrac{1}{{\tan C}}\\ \to \left( {\tan A + \tan B} \right)\tan C = 1 - \tan A.\tan B\\ \to \tan A.\tan B + \tan B.\tan C + \tan C.\tan A = 1\\ \to \tan {10^o}.\tan {25^o} + \tan {25^o}.\tan {55^o} + \tan {55^o}.\tan {10^o} = 1 \end{array}$
