Cho biết sin(alpha) x cos(alpha) = - 1/4 thì (tan ^2)alpha +(cot ^2)alpha bằng.

Ta có \[{\tan ^2}\alpha + {\cot ^2}\alpha = \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^4}\alpha + {{\cos }^4}\alpha }}{{{{\cos }^2}\alpha .{{\sin }^2}\alpha }} = \frac{{1 - 2{{\sin }^2}\alpha {{\cos }^2}\alpha }}{{{{\cos }^2}\alpha .{{\sin }^2}\alpha }} = \frac{{1 - 2.{{\left( { - \frac{1}{4}} \right)}^2}}}{{{{\left( { - \frac{1}{4}} \right)}^2}}} = 14\].

Chọn B.