Упрощение тригонометрических выражений. Задача 32math100admin44242025-03-24T14:13:59+03:00
Задача 32. Упростите выражение \(\dfrac{{{\text{ct}}{{\text{g}}^2}\left( {\alpha + \dfrac{\pi }{2}} \right)\,{{\cos }^2}\left( {\alpha -\dfrac{\pi }{2}} \right)}}{{{\text{ct}}{{\text{g}}^2}\left( {\alpha -\dfrac{\pi }{2}} \right)-{{\cos }^2}\left( {\alpha + \dfrac{\pi }{2}} \right)}}\)
Решение
\(\dfrac{{{\rm{ct}}{{\rm{g}}^2}\left( {\alpha + \dfrac{\pi }{2}} \right){{\cos }^2}\left( {\alpha -\dfrac{\pi }{2}} \right)}}{{{\rm{ct}}{{\rm{g}}^2}\left( {\alpha -\dfrac{\pi }{2}} \right)-{{\cos }^2}\left( {\alpha + \dfrac{\pi }{2}} \right)}} = \dfrac{{{\rm{t}}{{\rm{g}}^2}\alpha \cdot {{\sin }^2}\alpha }}{{{\rm{t}}{{\rm{g}}^2}\alpha -{{\sin }^2}\alpha }} = \)
\( = \dfrac{{\dfrac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} \cdot {{\sin }^2}\alpha }}{{\dfrac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}-{{\sin }^2}\alpha }} = \dfrac{{\dfrac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} \cdot {{\sin }^2}\alpha }}{{{{\sin }^2}\alpha \cdot \left( {\dfrac{1}{{{{\cos }^2}\alpha }}-1} \right)}} = \dfrac{{\dfrac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}}}{{\dfrac{{1-{{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }}}} = \dfrac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} \cdot \dfrac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = 1.\)
Ответ: 1.