Воспользуемся формулой суммы тангенсов:
\({\rm{tg}}\alpha {\rm{ + tg}}\beta {\rm{ = }}\dfrac{{\sin \left( {\alpha + \beta } \right)}}{{\cos \alpha \cos \beta }}.\)
\({\rm{tg}}{{\rm{9}}^ \circ } + {\rm{tg1}}{{\rm{5}}^ \circ }-{\rm{tg2}}{{\rm{7}}^ \circ }-{\rm{ctg2}}{{\rm{7}}^ \circ } + {\rm{ctg}}{{\rm{9}}^ \circ } + {\rm{ctg1}}{{\rm{5}}^ \circ } = \)
\( = {\rm{tg}}{{\rm{9}}^ \circ } + {\rm{ctg}}\left( {{{90}^ \circ }-{{81}^ \circ }} \right) + {\rm{tg1}}{{\rm{5}}^ \circ } + {\rm{ctg}}\left( {{{90}^ \circ }-{{75}^ \circ }} \right)-\left( {{\rm{tg2}}{{\rm{7}}^ \circ } + {\rm{ctg}}\left( {{{90}^ \circ }-{{63}^ \circ }} \right)} \right) = \)
\( = {\rm{tg}}{{\rm{9}}^ \circ } + {\rm{tg8}}{{\rm{1}}^ \circ } + {\rm{tg1}}{{\rm{5}}^ \circ } + {\rm{tg7}}{{\rm{5}}^ \circ }-\left( {{\rm{tg2}}{{\rm{7}}^ \circ } + {\rm{tg6}}{{\rm{3}}^ \circ }} \right) = \)
\( = \dfrac{{\sin {{90}^ \circ }}}{{\cos {9^ \circ }\cos {{81}^ \circ }}} + \dfrac{{\sin {{90}^ \circ }}}{{\cos {{15}^ \circ }\cos {{75}^ \circ }}}-\dfrac{{\sin {{90}^ \circ }}}{{\cos {{27}^ \circ }\cos {{63}^ \circ }}} = \)
\( = \dfrac{2}{{2\cos {9^ \circ }\sin {9^ \circ }}}-\dfrac{2}{{2\cos {{27}^ \circ }\sin {{27}^ \circ }}} + \dfrac{2}{{2\sin {{15}^ \circ }\cos {{15}^ \circ }}} = \)
\( = \dfrac{2}{{\sin {{18}^ \circ }}}-\dfrac{2}{{\sin {{54}^ \circ }}} + \dfrac{2}{{\sin {{30}^ \circ }}} = \dfrac{{2\sin {{54}^ \circ }-2\sin {{18}^ \circ }}}{{\sin {{18}^ \circ }\sin {{54}^ \circ }}} + 4 = \dfrac{{2\left( {\sin {{54}^ \circ }-\sin {{18}^ \circ }} \right)}}{{\sin {{18}^ \circ }\sin {{54}^ \circ }}} + 4 = \)
\( = \dfrac{{2 \cdot 2 \cdot \sin {{18}^ \circ } \cdot \cos {{36}^ \circ }}}{{\sin {{18}^ \circ } \cdot \sin \left( {{{90}^ \circ }-{{36}^ \circ }} \right)}} + 4 = \dfrac{{4\cos {{36}^ \circ }}}{{\cos {{36}^ \circ }}} + 4 = 4 + 4 = 8.\)
Ответ: 8.