Since ${\displaystyle \tan(x)={\frac {\sin(x)}{\cos(x)}}}$ , we can find its derivative by the usual rule for differentiating a fraction:

${\displaystyle {\frac {d}{dx}}\left[{\frac {\sin(x)}{\cos(x)}}\right]={\frac {\cos(x)\cdot \cos(x)+\sin(x)\cdot \sin(x)}{\cos ^{2}(x)}}={\frac {1}{\cos ^{2}(x)}}=\sec ^{2}(x)={1+\tan ^{2}(x)}}$ .

Similarly,

${\displaystyle {\frac {d}{dx}}{\bigl [}\cot(x){\bigr ]}=\csc ^{2}(x)=1+\cot ^{2}(x)}$

${\displaystyle {\frac {d}{dx}}{\bigl [}{\text{sec}}(x){\bigr ]}={\frac {\sin(x)}{\cos ^{2}(x)}}=\tan(x)\sec(x)}$

${\displaystyle {\frac {d}{dx}}{\bigl [}\csc(x){\bigr ]}=-{\frac {\cos(x)}{\sin ^{2}(x)}}=-\cot(x)\csc(x)}$