2 câu trả lời
Đáp án:
Giải thích các bước giải: \(\begin{array}{*{20}{l}}{\sin x = \frac{1}{5}}\\{*ta{\mkern 1mu} co:{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {{\sin }^2}x + \cos {{\mkern 1mu} ^2}x = 1}\\{ \Rightarrow cos{\mkern 1mu} x = \pm \sqrt {1 - {{\sin }^2}x} = \pm \sqrt {1 - {{\left( {\frac{1}{5}} \right)}^2}} = \pm \sqrt {1 - \frac{1}{{25}}} = \pm \sqrt {\frac{{24}}{{25}}} = \pm \frac{{2\sqrt 6 }}{5}}\\{ \Rightarrow cot{\mkern 1mu} x = \frac{{cosx}}{{\sin x}} = \frac{{ \pm \frac{{2\sqrt 6 }}{5}}}{{\frac{1}{5}}} = \pm 2\sqrt 6 }\\{ \Rightarrow \tan x = \frac{1}{{\cot {\mkern 1mu} x}} = \frac{1}{{ \pm 2\sqrt 6 }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( {vi{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \tan {\mkern 1mu} x.\cot {\mkern 1mu} x = 1} \right)}\end{array}\)
Ta có: ${\sin}^2x+{\cos }^2x=1$
$\sin x=\dfrac{1}{5}$
$\Rightarrow {\cos}^2x=1-{\sin}^2x$
$=1-\dfrac{1}{25}=\dfrac{24}{5}$
$\Rightarrow \cos x=\pm\dfrac{2\sqrt6}{5}$
$\Rightarrow \tan x=\dfrac{\sin x}{\cos x}=\dfrac{\pm1}{2\sqrt6}$
$\Rightarrow \cot x=\dfrac{1}{\tan x}=\pm2\sqrt6$