Cho góc \(\alpha \) thỏa điều kiện \(\frac{1}{{{{\tan }^2}\alpha }} + \frac{1}{{{{\cot }^2}\alpha }} + \frac{1}{{{{\sin }^2}\alpha }} + \frac{1}{{{{\cos }^2}\alpha }}{\mkern 1mu} {\mkern 1mu}  = 7\). Tính giá trị biểu thức \(P = \cos 4\alpha  + 2023.\) ?

Ta có

\(\begin{array}{l}\frac{1}{{{{\tan }^2}\alpha }} + \frac{1}{{{{\cot }^2}\alpha }} + \frac{1}{{{{\sin }^2}\alpha }} + \frac{1}{{{{\cos }^2}\alpha }}{\mkern 1mu} {\mkern 1mu}  = 7\\ \Leftrightarrow \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} + \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} + \frac{1}{{{{\sin }^2}\alpha }} + \frac{1}{{{{\cos }^2}\alpha }}{\mkern 1mu} {\mkern 1mu}  = 7\\ \Leftrightarrow \frac{{{{\cos }^4}\alpha  + {{\sin }^4}\alpha  + {{\sin }^2}\alpha  + {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha .{{\cos }^2}\alpha }} = 7\\ \Leftrightarrow \frac{{{{\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)}^2} - 2{{\sin }^2}\alpha .{{\cos }^2}\alpha  + 1}}{{{{\sin }^2}\alpha .{{\cos }^2}\alpha }} = 7\\ \Leftrightarrow 2 - 2{\sin ^2}\alpha .{\cos ^2}\alpha  - 7{\sin ^2}\alpha .{\cos ^2}\alpha  = 0\\ \Leftrightarrow {\sin ^2}\alpha .{\cos ^2}\alpha  = \frac{2}{9} \Leftrightarrow {\sin ^2}2\alpha  = \frac{8}{9}\end{array}\)

    Vậy \(P = \cos 4\alpha  + 2023 = 1 - 2{\sin ^2}2\alpha  + 2023 = 1 - 2.\frac{8}{9} + 2023 = \frac{{18200}}{9}\).