Упрощение тригонометрических выражений. Задача 11math100admin44242025-03-24T10:55:51+03:00
Задача 11. Докажите тождество \({\text{ct}}{{\text{g}}^2}\alpha -{\text{ct}}{{\text{g}}^2}\beta = \dfrac{{{{\cos }^2}\alpha -{{\cos }^2}\beta }}{{{{\sin }^2}\alpha \,{{\sin }^2}\beta }}\)
Решение
\({\rm{ct}}{{\rm{g}}^2}\alpha -{\rm{ct}}{{\rm{g}}^2}\beta = \dfrac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }}-\dfrac{{{{\cos }^2}\beta }}{{{{\sin }^2}\beta }} = \)
\( = \dfrac{{{{\cos }^2}\alpha {{\sin }^2}\beta -{{\cos }^2}\beta {{\sin }^2}\alpha }}{{{{\sin }^2}\alpha {{\sin }^2}\beta }} = \dfrac{{{{\cos }^2}\alpha \cdot \left( {1-{{\cos }^2}\beta } \right)-{{\cos }^2}\beta \cdot \left( {1-{{\cos }^2}\alpha } \right)}}{{{{\sin }^2}\alpha {{\sin }^2}\beta }} = \)
\( = \dfrac{{{{\cos }^2}\alpha -{{\cos }^2}\alpha {{\cos }^2}\beta -{{\cos }^2}\beta + {{\cos }^2}\beta {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha {{\sin }^2}\beta }} = \dfrac{{{{\cos }^2}\alpha -{{\cos }^2}\beta }}{{{{\sin }^2}\alpha {{\sin }^2}\beta }}.\)