Trigonometry/Derivative of Tangent

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

${\frac {d}{dx}}{\frac {\sin(x)}{\cos(x)}}={\frac {{\cos(x)\times \cos(x)}+{\sin(x)\times \sin(x)}}{\cos ^{2}(x)}}={\frac {1}{\cos ^{2}(x)}}=\sec ^{2}(x)=1+\tan ^{2}(x).$

Similarly,

$\displaystyle {\frac {d}{dx}}\cot(x)={\text{cosec}}^{2}(x)=1+\cot ^{2}(x).$
$\displaystyle {\frac {d}{dx}}{\text{sec}}(x)={\frac {\sin(x)}{\cos ^{2}(x)}}=\tan(x){\text{sec}}(x).$
$\displaystyle {\frac {d}{dx}}{\text{cosec}}(x)=-{\frac {\cos(x)}{\sin ^{2}(x)}}=-\cot(x){\text{cosec}}(x).$