Упрощение тригонометрических выражений. Задача 35

ОТВЕТ: \(2\,\left| {\,{\text{ctg}}\,\alpha \,} \right|\).

\(\sqrt {\left( {1-{\rm{t}}{{\rm{g}}^2}\dfrac{\alpha }{2}} \right)\left( {{\rm{ct}}{{\rm{g}}^2}\dfrac{\alpha }{2}-1} \right)}  = \sqrt {\left( {1-\dfrac{{{{\sin }^2}\dfrac{\alpha }{2}}}{{{{\cos }^2}\dfrac{\alpha }{2}}}} \right)\left( {\dfrac{{{{\cos }^2}\dfrac{\alpha }{2}}}{{{{\sin }^2}\dfrac{\alpha }{2}}}-1} \right)}  = \)

\( = \sqrt {\dfrac{{\left( {{{\cos }^2}\dfrac{\alpha }{2}-{{\sin }^2}\dfrac{\alpha }{2}} \right)\left( {{{\cos }^2}\dfrac{\alpha }{2}-{{\sin }^2}\dfrac{\alpha }{2}} \right)}}{{{{\cos }^2}\dfrac{\alpha }{2}{{\sin }^2}\dfrac{\alpha }{2}}}}  = \)

\( = \sqrt {\dfrac{{4\cos \alpha  \cdot \cos \alpha }}{{4{{\cos }^2}\dfrac{\alpha }{2}{{\sin }^2}\dfrac{\alpha }{2}}}}  = \sqrt {\dfrac{{4{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }}}  = \sqrt {4{\rm{ct}}{{\rm{g}}^2}\alpha }  = 2\left| {{\rm{ctg}}\alpha } \right|.\)

Ответ:  \(2\left| {{\rm{ctg}}\alpha } \right|.\)