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