Cho biết $\cos \alpha  + \sin \alpha  = \dfrac{1}{3}.$ Giá trị của $P = \sqrt {{{\tan }^2}\alpha  + {{\cot }^2}\alpha }$ bằng bao nhiêu ? 

Ta có $\cos \alpha  + \sin \alpha  = \dfrac{1}{3}$$\Rightarrow {\left( {\cos \alpha  + \sin \alpha } \right)^2} = \dfrac{1}{9}$

$\Leftrightarrow 1 + 2\sin \alpha \cos \alpha  = \dfrac{1}{9}$$\Leftrightarrow \sin \alpha \cos \alpha  =  - \dfrac{4}{9}$

Ta có $P = \sqrt {{{\tan }^2}\alpha  + {{\cot }^2}\alpha }$$= \sqrt {{{\left( {\tan \alpha  + \cot \alpha } \right)}^2} - 2\tan \alpha \cot \alpha }$ $= \sqrt {{{\left( {\dfrac{{\sin \alpha }}{{\cos \alpha }} + \dfrac{{\cos \alpha }}{{\sin \alpha }}} \right)}^2} - 2}$

$= \sqrt {{{\left( {\dfrac{{{{\sin }^2}\alpha  + {{\cos }^2}\alpha }}{{\sin \alpha \cos \alpha }}} \right)}^2} - 2}$ $= \sqrt {{{\left( {\dfrac{1}{{\sin \alpha \cos \alpha }}} \right)}^2} - 2}$ $= \sqrt {{{\left( { - \dfrac{9}{4}} \right)}^2} - 2}  = \dfrac{7}{4}$